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Contents, , Class XI, 1., , Some Basic Concepts of Chemistry ............................................... 1, , 2., , Structure of Atom ........................................................................ 8, , 3., , Classification of Elements and Periodicity in Properties ............. 16, , 4., , Chemical Bonding and Molecular Structure............................ 22, , 5., , States of Matter ......................................................................... 37, , 6., , Thermodynamics ....................................................................... 44, , 7., , Equilibrium ................................................................................ 54, , 8., , Redox Reactions ........................................................................ 71, , 9., , Hydrogen................................................................................... 74, , 10., , The s - Block Elements ................................................................ 77, , 11., , The p - Block Elements ............................................................... 83, , 12., , Organic Chemistry- Some Basic Principles and ....................... 87, Techniques, , www.neetujee.com, , 13., , Hydrocarbons ............................................................................ 97, , 14., , Environmental Chemistry.........................................................111, , www.mediit.in

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Class XII, , www.neetujee.com, , 1., , The Solid State ............................................................................. 1, , 2., , Solutions ..................................................................................... 8, , 3., , Electrochemistry........................................................................ 17, , 4., , Chemical Kinetics ....................................................................... 27, , 5., , Surface Chemistry...................................................................... 37, , 6., , General Principles and Processes of Isolation of Elements ............. 41, , 7., , The p - Block Elements................................................................ 44, , 8., , The d - and f - Block Elements ..................................................... 54, , 9., , Coordination Compounds ....................................................... 63, , 10., , Haloalkanes and Haloarenes ...................................................... 75, , 11., , Alcohols, Phenols and Ethers ..................................................... 84, , 12., , Aldehydes, Ketones and Carboxylic Acids ............................... 95, , 13., , Amines ....................................................................................110, , 14., , Biomolecules ...........................................................................119, , 15., , Polymers .................................................................................127, , 16., , Chemistry in Everyday Life .................................................... 132, , www.mediit.in

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Class XI, UNIT I, , , , , General Introduction: Important and scope of chemistry., Laws of chemical combination, Dalton’s atomic theory: concept of elements, atoms and molecules., Atomic and molecular masses. Mole concept and molar mass; percentage composition and, empirical and molecular formula; chemical reactions, stoichiometry and calculations based, on stoichiometry., , UNIT II, , , *, , Syllabus, , , , Chemical Bonding and Molecular Structure, , Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character, of covalent bond, valence bond theory, resonance, geometry of molecules, VSEPR theory, concept, of hybridization involving s, p and d orbitals and shapes of some simple molecules, molecular, orbital theory of homonuclear diatomic molecules (qualitative idea only). Hydrogen bond., , UNIT V, , , Classificationof Elementsand Periodicityin Properties, , Modern periodic law and long form of periodic table, periodic trends in properties of elements, atomic radii, ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity, valence., , UNIT IV, , , Structure of Atom, , Atomic number, isotopes and isobars. Concept of shells and subshells, dual nature of matter and, light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbital, quantum, numbers, shapes of s, p and d orbitals, rules for filling electrons in orbitals- Aufbau principle,, Pauli exclusion principles and Hund’s rule, electronic configuration of atoms, stability of half, filled and completely filled orbitals., , UNIT III, , , Some Basic Concepts of Chemistry, , States of Matter: Gases and Liquids, , Three states of matter, intermolecular interactions, types of bonding, melting and boiling, points, role of gas laws of elucidating the concept of the molecule, Boyle’s law, Charle’s law,, Gay Lussac’s law, Avogadro’s law, ideal behaviour of gases, empirical derivation of gas equation., Avogadro number, ideal gas equation. Kinetic energy and molecular speeds (elementary idea),, deviation from ideal behaviour, liquefaction of gases, critical temperature., Liquid State- Vapour pressure, viscosity and surface tension (qualitative idea only, no, mathematical derivations)., , UNIT VI Thermodynamics, , , , , , First law of thermodynamics-internal energy and enthalpy, heat capacity and specific, heat, measurement of U and H, Hess’s law of constant heat summation, enthalpy of : bond, dissociation, combustion, formation, atomization, sublimation, phase transition, ionization,, solution and dilution., Introduction of entropy as state function, Second law of thermodynamics, Gibbs energy change, for spontaneous and non-spontaneous process, criteria for equilibrium and spontaneity., Third law of thermodynamics- Brief introduction., , UNIT VII Equilibrium, , , Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of chemical, equilibrium, equilibrium constant, factors affecting equilibrium- Le Chatelier’s principle; ionic, equilibrium- ionization of acids and bases, strong and weak electrolytes, degree of ionization,, ionization of polybasic acids, acid strength, concept of pH., Hydrolysis of salts (elementary, idea)., buffer solutions, Henderson equation, solubility product, common ion effect (with, illustrative examples)., , UNIT VIII Redox Reactions, , , Concept of oxidation and oxidation and reduction, redox reactions oxidation number, balancing, redox reactions in terms of loss and gain of electron and change in oxidation numbers., , UNIT IX, , , Occurrence, isotopes, preparation, properties and uses of hydrogen; hydrides, ionic, covalent, and interstitial; physical and chemical properties of water, heavy water; hydrogen peroxidepreparation, reactions, uses and structure., , UNIT X, , , , Hydrogen, , s-Block Elements (Alkali and Alkaline Earth Metals), , Group I and group 2 elements:, General introduction, electronic configuration, occurrence, anomalous properties of the first, element of each group, diagonal relationship, trends in the variation of properties (such as, ionization enthalpy, atomic and ionic radii), trends in chemical reactivity with oxygen, water,, hydrogen and halogens; uses., *For details, refer to latest prospectus, , www.neetujee.com, , www.mediit.in

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, , , Preparation and Properties of Some important Compounds: Sodium carbonate, sodium chloride, sodium hydroxide and sodium hydrog encarbonate,, biological importance of sodium and potassium., Industrial use of lime and limestone, biological importance of Mg and Ca., , UNIT XI, , , , , , , Some p-Block Elements, , General Introduction to p-Block Elements., Group 13 elements: General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical, reactivity, anomalous properties of first element of the group; Boron, some important compounds: borax, boric acids, boron hydrides. Aluminium:, uses, reactions with acids and alkalies., General 14 elements: General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity,, anomalous behaviour of first element. Carbon, allotropic forms, physical and chemical properties: uses of some important compounds: oxides., Important compounds of silicon and a few uses: silicon tetrachloride, silicones, silicates and zeolites, their uses., , UNIT XII Organic Chemistry- Some Basic Principles and Techniques, , , , , General introduction, methods of purification qualitative and quantitative analysis, classification and IUPAC nomenclature of organic compounds., Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyperconjugation., Homolytic and heterolytic fission of a covalent bond: free radials, carbocations, carbanions; electrophiles and nucleophiles, types of organic reactions., , UNIT XIII Hydrocarbons, , , , , , , Alkanes- Nomenclature, isomerism, conformations (ethane only), physical properties, chemical reactions including free radical mechanism of, halogenation, combustion and pyrolysis., Alkenes-Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation: chemical reactions:, addition of hydrogen, halogen, water, hydrogen halides (Markovnikov’s addition and peroxide effect), ozonolysis, oxidation, mechanism of, electrophilic addition., Alkynes-Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of, alkynes, addition reaction of hydrogen, halogens, hydrogen halides and water., Aromatic hydrocarbons- Introduction, IUPAC nomenclature; Benzene; resonance, aromaticity; chemical properties: mechanism of electrophilic, substitution-Nitration sulphonation, halogenation, Friedel—Crafts alkylation and acylation; directive influence of functional group in monosubstituted benzene; carcinogenicity and toxicity., , UNIT XIV Environmental Chemistry, , , Environmental pollution: Air, water and soil pollution, chemical reactions in atmosphere, smogs, major atmospheric pollutants; acid rain ozone, and its reactions, effects of depletion of ozone layer, greenhouse effect and global warming-pollution due to industrial wastes; green chemistry as, an alternative tool for reducing pollution, strategy for control of environmental pollution., , Class XII, UNIT I, , , UNIT II, , , Surface Chemistry, , Adsorption-physisorption and chemisorption; factors affecting adsorption of gases on solids, catalysis homogeneous and heterogeneous, activity and, selectivity: enzyme catalysis; colloidal state: distinction between true solutions, colloids and suspensions; lyophillic, lyophobic multimolecular and, macromolecular colloids; properties of colloids; Tyndall effect, Brownian movement, electrophoresis, coagulation; emulsions- types of emulsions., , UNIT VI, , , Chemical Kinetics, , Rate of a reaction (average and instantaneous), factors affecting rates of reaction; concentration, temperature, catalyst; order and molecularity of, a reaction; rate law and specific rate constant, integrated rate equations and half life (only for zero and first order reactions); concept of collision, theory ( elementary idea, no mathematical treatment), Activation energy, Arrhenious equation., , UNIT V, , , Electrochemistry, , Redox reactions, conductance in electrolytic solutions, specific and molar conductivity variation of conductivity with concentration, Kohlrausch’s, Law, electrolysis and Laws of electrolysis (elementary idea), dry cell- electrolytic cells and Galvanic cells; lead accumulator, EMF of a cell, standard, electrode potential, Relation between Gibbs energy change and EMF of a cell, fuel cells; corrosion., , UNIT IV, , , Solutions, , Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids, solid solutions, colligative propertiesrelative lowering of vapour pressure, Raoult’s law, elevation of boiling point, depression of freezing point, osmotic pressur e, determination of, molecular masses using colligative properties abnormal molecular mass, van Hoff factor., , UNIT III, , , Solid State, , Classification of solids based on different binding forces; molecular, ionic, covalent and metallic solids, amorphous and crystalline solids, (elementary idea), unit cell in two dimensional and three dimensional lattices, calculation of density of unit cell, packing in solids, packing efficiency,, voids, number of atoms per unit cell in a cubic unit cell, point defects, electrical and magnetic properties, Band theory of metals, conductors,, semiconductors and insulators., , General Principles and Processes of Isolation of Elements, , Principles and methods of extraction- concentration, oxidation, reduction, electrolytic method and refining; occurrence and principles of extraction, of aluminium, copper, zinc and iron., , www.neetujee.com, , www.mediit.in

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UNIT VII p-Block Elements, , , , , , , , , Group 15 elements: General introduction, electronic configuration, occurrence, oxidation states, trends in physical and chemical properties;, preparation and properties of ammonia and nitric acid, oxides of nitrogen (structure only); Phosphorous- allotropic forms; compounds of, phosphorous: preparation and properties of phosphine, halides (PCl3, PCl5) and oxoacids (elementary idea only)., Group 16 elements: General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties; dioxygen:, preparation, properties and uses; classification of oxides; ozone. Sulphur – allotropic forms; compounds of sulphur: preparation, properties and, uses of sulphur dioxide; sulphuric acid: industrial process of manufacture, properties and uses, oxoacids of sulphur (structures only)., Group 17 elements: General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties;, compounds of halogens: preparation, properties and uses of chlorine and hydrochloric acid, interhalogen compounds oxoacids of halogens, (structures only)., Group 18 elements: General introduction, electronic configuration, occurrence, trends in physical and chemical properties, uses., , UNIT VIII d - and f - Block Elements, , , , , , General introduction, electronic configuration, characteristics of transition metals, general trends in properties of the first row transition metalsmetallic character, ionization enthalpy, oxidation states, ionic radii, colour, catalytic property, magnetic properties, interstitial compounds, alloy, formation. Preparation and properties of K2Cr2O7 and KMnO4., Lanthanoids- electronic configuration, oxidation states, chemical reactivity, and lanthanoid contraction and its consequences., Actinoids: Electronic configuration, oxidation states and comparison with lanthanoids., , UNIT IX, , , Coordination compounds: Introduction, ligands, coordination number, colour, magnetic properties and shapes, IUPAC nomenclature of mononuclear, coordination compounds, isomerism (structural and stereo) bonding, Werner’s theory VBT,CFT; importance of coordination compounds (in, qualitative analysis, biological systems)., , UNIT X, , , , , , , , Haloalkanes and Haloarenes, , Haloalkanes: Nomenclature, nature of C—X bond, physical and chemical properties, mechanism of substitution reactions. Optical rotation., Haloarenes: Nature of C—X bond, substitution reactions (directive influence of halogen for monosubstituted compounds only)., Uses and environment effects of – dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons, DDT., , UNIT XI, , , Coordination Compounds, , Alcohols, Phenols and Ethers, , Alcohols: Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only); identification of primary, secondary, and tertiary alcohols; mechanism of dehydration, uses with special reference to methanol and ethanol., Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol, electrophillic substitution reactions,, uses of phenols., Ethers: Nomenclature, methods of preparation, physical and chemical properties uses., , UNIT XII Aldehydes, Ketones and Carboxylic Acids, , , , Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties; and mechanism of, nucleophilic addition, reactivity of alpha hydrogen in aldehydes; uses., Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses., , UNIT XIII Organic Compounds Containing Nitrogen, , , , , Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses, identificati on of primary secondary, and tertiary amines., Cyanides and Isocyanides- will be mentioned at relevant places., Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry., , UNIT XIV Biomolecules, , , , , , , Carbohydrates- Classification (aldoses and ketoses), monosaccharide (glucose and fructose), D.L. configuration, oligosaccharides (sucrose, lactose,, maltose), polysaccharides (starch, cellulose, glycogen): importance., Proteins- Elementary idea of – amino acids, peptide bond, polypeptides, proteins, primary structure, secondary structure, tertiary structure and, quaternary structure (qualitative idea only), denaturation of proteins; enzymes., Hormones- Elementary idea (excluding structure)., Vitamins- Classification and function., Nucleic Acids: DNA and RNA, , UNIT XV Polymers, , , Classification- Natural and synthetic, methods of polymerization (addition and condensation), copolymerization. Some important polymers:, natural and synthetic like polyesters, bakelite; rubber, Biodegradable and non-biodegradable polymers., , UNIT XVI Chemistry in Everyday Life, , , , , Chemicals in medicines- analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids, antihistamines., Chemicals in food- preservatives, artificial sweetening agents, elementary idea of antioxidants., Cleansing agents- soaps and detergents, cleansing action., , , www.neetujee.com, , www.mediit.in

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, , , , , , , CHAPTER, , 1, , 1.3, 1., , Uncertainty in Measurement, , Given the numbers : 161 cm, 0.161 cm, 0.0161 cm. The, number of significant figures for the three numbers, is, (a) 3, 3 and 4 respectively, (b) 3, 4 and 4 respectively, (c) 3, 4 and 5 respectively, (d) 3, 3 and3 respectively., (1998), , 1.5 Laws of Chemical Combinations, 3., , (a) N/2, (c) 2 N, , Properties of Matter and their Measurement, , The dimensions of pressure are the same as that of, (a) force per unit volume, (b) energy per unit volume, (c) force, (d) energy, (1995), , 1.4, 2., , Some Basic Concepts, of Chemistry, , Equal masses of H2, O2 and methane, have been taken in a container of volume V at, temperature 27 °C in identical conditions. The ratio, of the volumes of gases H2 : O2 : methane would be, (a) 8 : 16 : 1, (b) 16 : 8 : 1, (c) 16 : 1 : 2, (d) 8 : 1 : 2, (2014), , 4., , What volume of oxygen gas (O2) measured at, 0°C and 1 atm, is needed to burn completely 1 L, of propane gas (C3H8) measured under the same, conditions?, (a) 5 L, (b) 10 L, (c) 7 L, (d) 6 L, (2008), , 5., , 0.24 g of a volatile gas, upon vaporisation, gives 45 mL, vapour at NTP. What will be the vapour density of, the substance? (Density of H2 = 0.089 g/L), (a) 95.93, (b) 59.93, (c) 95.39, (d) 5.993, (1996), , 6., , The molecular weight of O2 and SO2 are 32 and, 64 respectively. At 15°C and 150 mmHg pressure,, one litre of O 2 contains ‘N’ molecules. The number, of molecules in two litres of SO2 under the same, conditions of temperature and pressure will be, , www.neetujee.com, , 7., , (b) N, (d) 4 N, , (1990), , What is the weight of oxygen required for the, complete combustion of 2.8 kg of ethylene?, (a) 2.8 kg, (b) 6.4 kg, (c) 9.6 kg, (d) 96 kg, (1989), , 1.7 Atomic and Molecular Masses, 8., , An element, X has the following isotopic composition :, 200, 202, X : 90% 199X : 8.0%, X : 2.0%, The weighted average atomic mass of the naturally, occurring element X is closest to, (a) 201 amu, (b) 202 amu, (c) 199 amu, (d) 200 amu, (2007), , 9., , Boron has two stable isotopes, 10B(19%) and, 11, B(81%). Calculate average at. wt. of boron in the, periodic table., (a) 10.8, (b) 10.2, (c) 11.2, (d) 10.0, (1990), , 1.8 Mole Concept and Molar Masses, 10. Which one of the followings has maximum number, of atoms?, (a) 1 g of Ag(s) [Atomic mass of Ag = 108], (b) 1 g of Mg(s) [Atomic mass of Mg = 24], (c) 1 g of O2(g) [Atomic mass of O = 16], (d) 1 g of Li(s) [Atomic mass of Li = 7], (NEET 2020), 11. In which case is number of molecules of water, maximum?, (a) 18 mL of water, (b) 0.18 g of water, (c) 0.00224 L of water vapours at 1 atm and 273 K, (d) 10–3 mol of water, (NEET 2018), 12. Suppose the elements X and Y combine to form two, compounds XY2 and X3Y2. When 0.1 mole of XY2, weighs 10 g and 0.05 mole of X3Y2 weighs 9 g, the, atomic weights of X and Y are, , www.mediit.in

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2, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (a) 40, 30, (c) 20, 30, , (b) 60, 40, (d) 30, 20 (NEET-II 2016), , 13. If Avogadro number NA, is changed from, 6.022 × 1023 mol–1 to 6.022 × 1020 mol–1, this would, change, (a) the mass of one mole of carbon, (b) the ratio of chemical species to each other in a, balanced equation, (c) the ratio of elements to each other in a, compound, (d) the definition of mass in units of grams. (2015), 14. The number of water molecules is maximum in, (a) 1.8 gram of water, (b) 18 gram of water, (c) 18 moles of water, (d) 18 molecules of water., (2015), 15. A mixture of gases contains H2 and O2 gases in the, ratio of 1 : 4 (w/w). What is the molar ratio of the, two gases in the mixture?, (a) 16 : 1, (b) 2 : 1, (c) 1 : 4, (d) 4 : 1 (2015, Cancelled), 16. Which has the maximum number of molecules, among the following?, (a) 44 g CO2, (b) 48 g O3, (c) 8 g H2, (d) 64 g SO2, (Mains 2011), 17. The number of atoms in 0.1 mol of a triatomic gas is, (NA = 6.02 × 1023 mol–1), (a) 6.026 × 1022, (b) 1.806 × 1023, (c) 3.600 × 1023, (d) 1.800 × 1022, (2010), 18. The maximum number of molecules is present in, (a) 15 L of H2 gas at STP, (b) 5 L of N2 gas at STP, (c) 0.5 g of H2 gas, (d) 10 g of O2 gas., (2004), 19. Which has maximum molecules?, (a) 7 g N2, (b) 2 g H2, (c) 16 g NO2, (d) 16 g O2, , (2002), , 20. Specific volume of cylindrical virus particle is, 6.02 × 10–2 cc/g whose radius and length are 7 Å and, 10 Å respectively. If NA = 6.02 × 1023, find molecular, weight of virus., (a) 15.4 kg/mol, (b) 1.54 × 104 kg/mol, 4, (c) 3.08 × 10 kg/mol (d) 3.08 × 103 kg/mol, (2001), 21. The number of atoms in 4.25 g of NH3 is, approximately, (a) 4 × 1023, (b) 2 × 1023, 23, (c) 1 × 10, (d) 6 × 1023, , www.neetujee.com, , (1999), , 22. Haemoglobin contains 0.334% of iron by weight. The, molecular weight of haemoglobin is approximately, 67200. The number of iron atoms (Atomic weight of, Fe is 56) present in one molecule of haemoglobin is, (a) 4, (b) 6, (c) 3, (d) 2, (1998), 23. The number of moles of oxygen in one litre of air, containing 21% oxygen by volume, under standard, conditions, is, (a) 0.0093 mol, (b) 2.10 mol, (c) 0.186 mol, (d) 0.21 mol, (1995), 24. The total number of valence electrons in 4.2 g of N3–, ion is (NA is the Avogadro’s number), (a) 2.1 NA, (b) 4.2 NA, (c) 1.6 NA, (d) 3.2 NA, (1994), 25. The number of gram molecules of oxygen in, 6.02 × 1024 CO molecules is, (a) 10 g molecules, (b) 5 g molecules, (c) 1 g molecule, (d) 0.5 g molecules., (1990), 26. Ratio of Cp and Cv of a gas ‘X’ is 1.4. The number, of atoms of the gas ‘X’ present in 11.2 litres of it at, NTP will be, (a) 6.02 × 1023, (b) 1.2 × 1023, 23, (c) 3.01 × 10, (d) 2.01 × 1023, (1989), 27. The number of oxygen atoms in 4.4 g of CO2 is, (a) 1.2 × 1023, (b) 6 × 1022, (c) 6 × 1023, (d) 12 × 1023, (1989), 28. 1 cc N2O at NTP contains, 1.8, 1022 atoms, 224, 6.02, (b), 1023 molecules, 22400, 1.32, (c), 1023 electrons, 224, (d) all of the above., (a), , 1.9, , (1988), , Percentage Composition, , 29. An organic compound contains carbon, hydrogen, and oxygen. Its elemental analysis gave C, 38.71%, and H, 9.67%. The empirical formula of the, compound would be, (a) CHO, (b) CH 4O, (c) CH3O, (d) CH2O, (2008), 30. Percentage of Se in peroxidase anhydrous enzyme, is 0.5% by weight (at. wt. = 78.4) then minimum, molecular weight of peroxidase anhydrous enzyme is, (a) 1.568 × 104, (b) 1.568 × 103, (c) 15.68, (d) 2.136 × 104, (2001), , www.mediit.in

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Some Basic Concepts of Chemistry, 31. Which of the following fertilizers has the highest, nitrogen percentage?, (a) Ammonium sulphate, (b) Calcium cyanamide, (c) Urea, (d) Ammonium nitrate, (1993), , 1.10 Stoichiometry and Stoichiometric, Calculations, 32. The number of moles of hydrogen molecules, required to produce 20 moles of ammonia through, Haber’s process is, (a) 40, (b) 10, (c) 20, (d) 30, (NEET 2019), 33. The density of 2 M aqueous solution of NaOH is, 1.28 g/cm3. The molality of the solution is [Given, that molecular mass of NaOH = 40 g mol–1], (a) 1.20 m, (b) 1.56 m, (c) 1.67 m, (d) 1.32 m, (Odisha NEET 2019), 34. A mixture of 2.3 g formic acid and 4.5 g oxalic acid, is treated with conc. H2SO4. The evolved gaseous, mixture is passed through KOH pellets. Weight (in g), of the remaining product at STP will be, (a) 1.4, (b) 3.0, (c) 2.8, (d) 4.4 (NEET 2018), 35. What is the mass of the precipitate formed when, 50 mL of 16.9% solution of AgNO3 is mixed with, 50 mL of 5.8% NaCl solution?, (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5), (a) 3.5 g, (b) 7 g, (c) 14 g, (d) 28 g, (2015), 36. 20.0 g of a magnesium carbonate sample, decomposes on heating to give carbon dioxide and, 8.0 g magnesium oxide. What will be the percentage, purity of magnesium carbonate in the sample?, (At. wt. of Mg = 24), (a) 96, (b) 60, (c) 84, (d) 75, (2015), 37. When 22.4 litres of H2(g) is mixed with 11.2 litres, of Cl2(g), each at STP, the moles of HCl(g) formed is, equal to, (a) 1 mol of HCl(g) (b) 2 mol of HCl(g), (c) 0.5 mol of HCl(g) (d) 1.5 mol of HCl(g), (2014), 38. 1.0 g of magnesium is burnt with 0.56 g O2 in a, closed vessel. Which reactant is left in excess and, how much? (At. wt. Mg = 24, O = 16), , www.neetujee.com, , 3, , (a) Mg, 0.16 g, (c) Mg, 0.44 g, , (b) O2, 0.16 g, (d) O2, 0.28 g, , (2014), , 20, , 39. 6.02 × 10 molecules of urea are present in 100 mL, of its solution. The concentration of solution is, (a) 0.001 M, (b) 0.1 M, (c) 0.02 M, (d) 0.01 M, (NEET 2013), 40. In an experiment it showed that 10 mL of 0.05 M, solution of chloride required 10 mL of 0.1 M, solution of AgNO3, which of the following will be, the formula of the chloride (X stands for the symbol, of the element other than chlorine)?, (a) X2Cl2, (b) XCl2, (c) XCl4, (d) X2Cl, (Karnataka NEET 2013), 41. 25.3 g of sodium carbonate, Na2CO3 is dissolved in, enough water to make 250 mL of solution. If sodium, carbonate dissociates completely, molar concentration, of sodium ion, Na+ and carbonate ions, CO32– are, respectively, (Molar mass of Na2CO3 = 106 g mol–1), (a) 0.955 M and 1.910 M, (b) 1.910 M and 0.955 M, (c) 1.90 M and 1.910 M, (d) 0.477 M and 0.477 M, (2010), 42. 10 g of hydrogen and 64 g of oxygen were filled in a, steel vessel and exploded. Amount of water produced, in this reaction will be, (a) 3 mol, (b) 4 mol, (c) 1 mol, (d) 2 mol, (2009), 43. How many moles of lead(II) chloride will be formed, from a reaction between 6.5 g of PbO and 3.2 g HCl?, (a) 0.011, (b) 0.029, (c) 0.044, (d) 0.333, (2008), 44. The mass of carbon anode consumed (giving only, carbon dioxide) in the production of 270 kg of, aluminium metal from bauxite by the Hall process is, (a) 270 kg, (b) 540 kg, (c) 90 kg, (d) 180 kg, (Atomic mass : Al = 27), (2005), 45. Molarity of liquid HCl, if density of solution is, 1.17 g/cc is, (a) 36.5, (b) 18.25, (c) 32.05, (d) 42.10, (2001), 46. Volume of CO2 obtained by the complete, decomposition of 9.85 g of BaCO3 is, (a) 2.24 L, (b) 1.12 L, (c) 0.84 L, (d) 0.56 L, , (2000), , www.mediit.in

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4, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 47. In the reaction,, 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l), when 1 mole of ammonia and 1 mole of O2 are made, to react to completion, (a) all the oxygen will be consumed, (b) 1.0 mole of NO will be produced, (c) 1.0 mole of H 2O is produced, (d) all the ammonia will be consumed., (1998), , 48. The amount of zinc required to produce 224 mL of, H2 at STP on treatment with dilute H2SO4 will be, (a) 65 g, (b) 0.065 g, (c) 0.65 g, (d) 6.5 g, (1996), 49. At STP the density of CCl4 vapour in g/L will be, nearest to, (a) 6.87, (b) 3.42, (c) 10.26, (d) 4.57, (1988), , ANSWER KEY, , 1., 11., 21., 31., 41., , (b), (a), (d), (c), (b), , 2., 12., 22., 32., 42., , 3., 13., 23., 33., 43., , (d), (a), (a), (d), (b), , (c), (a), (a), (c), (b), , 4., 14., 24., 34., 44., , (a), (c), (c), (c), (c), , 5., 15., 25., 35., 45., , (b), (d), (b), (b), (c), , 6., 16., 26., 36., 46., , (c), (c), (a), (c), (b), , 7., 17., 27., 37., 47., , (c), (b), (a), (a), (a), , 8., 18., 28., 38., 48., , (d), (a), (d), (a), (c), , 9., 19., 29., 39., 49., , (a), (b), (c), (d), (a), , 10., 20., 30., 40., , (d), (a), (a), (b), , Hints & Explanations, 1., , (b) : Pressure , , Force, , , , Area, MLT2, Therefore, dimensions of pressure , = ML–1T–2, 2, L, and dimensions of energy per unit volume, Energy ML2T2,  ML1 T2, , , Volume, L3, 2. (d) : Zeros placed left to the number are never, significant, therefore the no. of significant figures for the, numbers 161 cm, 0.161 cm and 0.0161 cm are same, i.e., 3., 3. (c) : According to Avogadro’s hypothesis, ratio of, the volumes of gases will be equal to the ratio of their no., of moles., Mass, So, no. of moles =, Mol. mass, w, w, nH  ; nO  ; n  w, 2, 2, 2, 32 CH 4 16, w w, So, the ratio is :, w or 16 :1: 2., :, 2 32 16, 4. (a) : C3H8 + 5O2, 3CO2 + 4H2O, 1 vol., , 5 vol., , 3 vol., , 4 vol., , According to the above equation,, 1 vol. or 1 litre of propane requires 5 vol. or 5 litres of O2, to burn completely., 5. (b) : Weight of gas = 0.24 g,, Volume of gas = 45 mL = 0.045 litre and density of, H2 = 0.089 g/L, Weight of 45 mL of H2 = density × volume, = 0.089 × 0.045 = 4.005 × 10–3 g, , www.neetujee.com, , Therefore, vapour density, Weight of certain volume of substance, , Weight of same volume of hydrogen, 0.24, ,  59.93, 4.005 103, 6. (c) : If 1 L of one gas contains N molecules, 2 L of any, gas under the same conditions will contain 2N molecules., 7. (c) : C2H4 + 3O2  2CO2 + 2H2O, 28 g, , 96 g, , For complete combustion,, 96, 2.8 kg of C2H4 requires   2.8 103 g, 28, 3, , = 9.6 × 10 g = 9.6 kg of O2, 8., , (d) : Average isotopic mass of X, 200  90 199  8  202  2, , 90  8  2, 18000 1592  404, ,  199.96 amu  200 amu, 100, 19 10  81 11, 9. (a) : Average atomicmass ,  10.81, 100, 10. (d) : 1 mole of substance = NA atoms, NA, atoms, 108 g of Ag = NA atoms  1g of Ag , 108, NA, atoms, 24 g of Mg = NA atoms  1g of Mg , 24, 32 g of O2 = NA molecules = 2 NA atoms, N,  1g of O  A atoms, 2, 16, , www.mediit.in

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5, , Some Basic Concepts of Chemistry, 7 g of Li = NA atoms  1g of Li , , NA, , atoms, , 7, Therefore, 1 g of Li(s), has maximum number of atoms., 11. (a) : (a) Mass of water = V × d = 18 × 1 = 18 g, 18, Molecules of water = mole × NA  NA  N A, 18, 0.18, (b) Molecules of water = mole × NA  18 NA, = 10–2 NA, 0.00224, (c) Moles of water ,  104, 22.4, –4, Molecules of water = mole × NA = 10 NA, (d) Molecules of water = mole × NA = 10–3 NA, 12. (a) : Let atomic weight of element X is x and that of, element Y isw, y., For XY , n , 2, , 0.1 , , 10, x  2y, , Mol. wt.,  x + 2y =, , 10, 0.1, ,  100, , ...(i), , 12  6.022 1020,  12 103 g, 6.022 1023, 6.023 1023, 14. (c) : 1.8 gram of water , 1.8, 18, = 6.023 × 1022 molecules, 18 gram of water = 6.023 × 1023 molecules, 18 moles of water = 18 × 6.023 × 1023 molecules, 15. (d) : Number of moles of H2 = 1/2, 4, Number of moles of O2 , 32, 1 4, Hence, molar ratio  :  4 : 1, 2 32, 16. (c) :, 8 g H2 has 4 moles while the others has 1 mole, each., , , of moles × 3, 17. (b) : No. of atoms = NA × No., 23, 23, = 6.023 × 10 × 0.1 × 3 = 1.806 × 10, 18. (a) : At STP, 22.4 L = 6.023 × 1023 molecules, 6.023 1023 15, ,  4.033 1023 molecules, 15 L H, , www.neetujee.com, , 22.4, , 0.5 g H2 , , 6.023 1023  0.5,  1.505 1023 molecules, 2, , 32 g O2 = 6.023 × 1023 molecules, 6.023 1023 10, 23, 10 g of O2 , molecules, , 1.882, 10, 32, 19. (b) : Number of molecules = moles × NA, 7, Molecules of N2 = NA = 0.5 NA, 14, Molecules of H2 = NA, 16, NA = 0.35 NA, 46, 16, Molecules of O2 =, N = 0.5 NA, 32, A,  2 g H2 (1 mole H2) contains maximum molecules., 20. (a) : Specific volume (vol. of 1 g) of cylindrical virus, particle = 6.02 × 10–2 cc/g, Radius of virus, r = 7 Å = 7 × 10–8 cm, Volume of virus = r2l, 22,   (7 108 )2 10 108 = 154 × 10–23 cc, 7, Volume (cc), wt. of one virus particle , Specific volume (cc/g), Molecules of NO2 =, , w, For X3Y2, n , Mol. wt., 9, 9, 0.05 ,  3x  2 y   180, ...(ii), 3x  2 y, 0.05, On solving equations (i) and (ii), we get x = 40, 40 + 2y = 100  2y = 60  y = 30, 13. (a) : Mass of 1 mol (6.022 × 1023 atoms) of carbon, = 12 g, 20, If Avogadro number is changed to 6.022 × 10 atoms, then mass of 1 mol of carbon, , 2, , 6.023 1023  5,  1.344 1023 molecules, 22.4, 2 g H2 = 6.023 × 1023 molecules, 5 L N2 , , 154 1023,  6.02 102 g, , , Molecular wt. of virus = wt. of N particles, , , 154 10, , 23, 2, , A, ,  6.02 1023 g/mol, , 6.02 10, = 15400 g/mol = 15.4 kg/mol, 21. (d) : 17 g of NH3 = 4NA atoms, 4N A,  4.25 atoms, 4.25 g of NH , 3, 17, = NA atoms = 6 × 1023 atoms, 22. (a) : Quantity of iron in one molecule, 67200, =, × 0.334 = 224.45 amu, 100, No. of iron atoms in one molecule of haemoglobin, 224.45, , 4, 56, 23. (a) : Volume of oxygen in one litre of air, 21,  1000  210 mL, 100, 210, Therefore, no. of moles =, , 22400, , = 0.0093 mol, , www.mediit.in

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6, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 24. (c) : Each nitrogen atom has 5 valence electrons,, therefore total number of valence electrons in N3– ion, is 16. Since the molecular mass of N– 3 is 42, therefore, total number of valence electrons in 4.2 g of N– ion3, 4.2, =, 16  N  1.6 N, A, A, 42, 25. (b) : Avogadro’s no., NA = 6.02 × 1023 molecules = 1 mole,  6.02 × 1024 CO molecules = 10 moles CO, = 10 g atoms of O = 5 g molecules of O2, 26. (a) : Here, Cp/Cv = 1.4, which shows that the gas is, diatomic., 22.4 L at NTP = 6.02 × 1023 molecules,  11.2 L at NTP = 3.01 × 1023 molecules, Since gas is diatomic,,  11.2 L at NTP = 2 × 3.01 × 1023 atoms, = 6.02 × 1023 atom, 27. (a) : 1 mol of CO2 = 44 g of CO2, , , 4.4 g CO2 = 0.1 mol CO2 = 6 ×23 1022 molecules, [Since, 1 mole CO2 = 6 × 10 molecules], = 2 × 6 × 1022 atoms of O = 1.2 × 1023 atoms of O, , 28. (d) : As we know,, 22400 cc of N2O contain 6.02 × 1023 molecules, , , 78.4 g Se will be present in, , 0.5, Minimum molecular weight of enzyme is, 1.568 × 104., 28, 31. (c) : Urea (NH2CONH2), % of N   100  46.66%, 60, Similarly, % of N in other compounds are :, (NH4)2SO4 = 21.2%; CaCN2 = 35.0% and NH4NO3 = 35.0%, 32. (d) : Haber’s process, N2 + 3H2  2NH3, 2 moles of NH3 are formed by 3 moles of H2.,  20 moles of NH3 will be formed by 30 moles of H2., 33. (c) : Density = 1.28 g/cc,, Conc. of solution = 2 M, Molar mass of NaOH = 40 g mol–1, Volume of solution = 1 L = 1000 mL, Mass of solution = d × V = 1.28 × 1000 = 1280 g, Mass of solute = n × Molar mass = 2 × 40 = 80 g, Mass of solvent = (1280 – 80) g = 1200 g, 80, Number of moles of solute =, =2, 40, 2 1000,  Molality =,  1.67 m, 1200, D ehydrating agent, 34. (c) : HCOOH        CO + H O, , Atomic, mass, , C, , 38.71, , 12, , H, , 9.67, , 1, , Mole, , O, , 51.62, , 16, , Simple, ratio, ,  3.22 3.22  1, 3.22, 12,  9.67, , 1, , 9.67, 3, 3.22, , 51.62, ,  3.22 3.22  1, 3.22, 16, , Hence, empirical formula of the compound would be, CH3O., 30. (a) : In peroxidase anhydrous enzyme, 0.5% Se is, present means, 0.5 g Se is present in 100 g of enzyme. In a, molecule of enzyme one Se atom must be present. Hence,, , www.neetujee.com, , 2.3 1, , 46 20, nf = 0, ni , , 38.71, , 9.67, , 2, , conc. H2SO4, , Since in N2O molecule there are 3 atoms, 1.8 1022, 3  6.02 1023,  1 cc N O , atoms , atoms, 2, 224, 22400, No. of electrons in a molecule of N2O = 7 + 7 + 8 = 22, Hence, no. of electrons in 1 cc of N2O, 6.02 1023, 1.32, ,  22 electrons , 1023 electrons, 22400, 224, 29. (c) :, %, ,  78.4  1.568  104, , , , 1023, 1 cc of N2O contain 6.02, 22400 molecules, , Element, , 100, , conc. H SO, , , , 0, , 0, , 1, 20, , 1, 20, , 4, H2C2 4 O   2 , CO + CO 2 + H2O, , ni , nf =0, , 4.5 1, , 90 20, , 0, , 0, , 0, , 1, 20, , 1, 20, , 1, 20, , H2O gets absorbed by conc. H2SO4. Gaseous mixture, (containing CO and CO2) when passed through KOH, pellets, CO2 gets absorbed., 1 1 1, Moles of CO left (unabsorbed)   , 20 20 10, 1, Mass of CO = moles × molar mass   28  2.8 g, 10, 35. (b) : 16.9% solution of AgNO3 means 16.9 g of AgNO3, in 100 mL of solution., = 8.45 g of AgNO3 in 50 mL solution., Similarly, 5.8 g of NaCl in 100 mL solution,  2.9 g of NaCl in 50 mL solution., The reaction can be represented as :, AgNO3 + NaCl, AgCl + NaNO3, Initial, 8.45/170 2.9/58.5, = 0.049, = 0.049, mole, Final moles 0, 0, , , , 0, , 0, , 0.049, , 0.049, , Mass of AgCl precipitated = 0.049 × 143.3, = 7.02  7 g, , www.mediit.in

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7, , Some Basic Concepts of Chemistry, ,  MgO (s) + CO2(g), 36. (c) : MgCO3(s) , , 84 g, ,  In this reaction, oxygen is the limiting reagent so, amount of H2O produced depends on the amount of O2., Since 0.5 mol of O2 gives 1 mol of H2O,  2 mol of O2 will give 4 mol of H2O, , 40 g, , 84 g of MgCO3  40 g of MgO, 40,  20 g of MgCO3   20 = 9.52 g of MgO, 84, Actual yield = 8 g of MgO, 8,  % purity = 100 = 84%, 9.52, 37. (a) : 1 mole  22.4 litres at STP., 11.2, 22.4,  0.5 mol,  1 mol; n , nH2 , Cl2, 22.4, 22.4, Reaction is as,, H2(g) +, Cl2(g), 2HCl(g), Initial, Final, , 1 mol, (1 – 0.5), = 0.5 mol, , 0.5 mol, (0.5 – 0.5), = 0 mol, , 43. (b) : PbO + 2HCl  PbCl2 + H2O, 6.5, 224, , (From bauxite), , 0, 2 × 0.0175, , Here, O2 is limiting reagent.,  Mass of Mg left in excess = 0.0066 × 24 = 0.16 g, 6.02 1020, , 3, , 2, , 3, , 2g, 1 mol, , 16 g, 0.5 mol, , 18 g, 1 mol, , 10 g of H2 = 5 mol and 64 g of O2 = 2 mol, , www.neetujee.com, , 4 moles of Al is produced by 3 moles of C., 1 mole of Al is produced by 3 mole of C., 4, 270  1000, 3, 4, 4, × 10, = 10 moles of Al is produced by, 4, 27, moles of C., 3, 4, Amount of carbon used = × 10 × 12 g, 4, 3, = × 10 × 12 kg = 90 kg, 4, 45. (c) : Density = 1.17 g/cc.,  1 cc. solution contains 1.17 g of HCl, 1.17 1000,  Molarity = 36.5 1 = 32.05, 46. (b) : BaCO, 197.3 g3  BaO + CO2, ,  0.001, 6.02 1023, 0.001, Concentration of solution ,  1000  0.01 M, 100, 40. (b) : Millimoles of solution of chloride, = 0.05 × 10 = 0.5, Millimoles of AgNO3 solution = 10 × 0.1 = 1, So, the millimoles of AgNO3 are double than the chloride, solution.,  XCl2 + 2AgNO3  2AgCl + X(NO3)2, 41. (b) : Given that molar mass of Na2CO3 = 106 g, 25.3  1000,  Molarity of solution , = 0.955 M, 106  250, Na2CO3  2Na+ + CO32–, [Na+]2–= 2[Na2CO3] = 2 × 0.955 = 1.910 M, [CO ] = [Na CO ] = 0.955 M, 42. (b) : H2 + 1/2O2  H2O, , mol, , 36.5, , = 0.029 mol = 0.087 mol, , 0, 2 × 0.5, 1 mol, , 39. (d) : Moles of urea , , 3.2, , Formation of moles of lead(II) chloride depends upon, the no. of moles of PbO which acts as a limiting reagent, here. So, no. of moles of PbCl2 formed will be equal to the, no. of moles of PbO i.e. 0.029., 44. (c) : 3C + 2Al2O3, 4Al + 3CO2, , Here, Cl2 is limiting reagent. So, 1 mole of HCl(g) is formed., 1, 38. (a) : nMg  = 0.0416 moles, 24, 0.56, = 0.0175 mole, nO2 , 32, The balanced equation is, 2Mg, +, O2, 2MgO, Initial 0.0416 mole, 0.0175 mole, Final (0.0416 – 2 × 0.0175), 0, = 0.0066 mole, , mol, , 22.4 L at N.T.P., 22.4, × 9.85, 197.3, = 1.118 L, , 9.85 g, ,  9.85 g of BaCO3 will produce 1.118 L of CO2 at, N.T.P. on the complete decomposition., 47. (a) : 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l), 4 moles, , 5 moles, , 4 moles, , 6 moles, ,  1 mole of NH3 requires = 5/4 = 1.25 mole of oxygen, while 1 mole of O2 requires =4/5 = 0.8 mole of NH3., Therefore, all oxygen will be consumed., 48. (c) : Zn + H2SO4  ZnSO4 + H2, 65 g, , 22400 mL, , Since 65 g of zinc reacts to liberate 22400 mL of H2 at, STP, therefore amount of zinc needed to produce 224 mL, 65, of H2 at STP   224  0.65 g, 22400, , 49. (a) : Weight of 1 mol of CCl4 vapour, = 12 + 4 × 35.5 = 154 g, 154, g L1 = 6.875 g L–1,  Density of CCl4 vapour , 22.4, , www.mediit.in

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CHAPTER, , 2, , 2.2, 1., , Structure of Atom, , Atomic Models, , The number of protons, neutrons and electrons in, 71 Lu, respectively, are, (a) 71, 104 and 71, (b) 104, 71 and 71, (c) 71, 71 and 104, (d) 175, 104 and 71, (NEET 2020), , 8., , The value of Planck’s constant is 6.63 × 10–34 Js. The, speed of light is 3 × 1017 nm s–1. Which value is, closest to the wavelength in nanometer of a quantum, of light with frequency of 6 × 1015 s–1?, (a) 50, (b) 75, (c) 10, (d) 25, (NEET 2013), , 9., , According to law of photochemical equivalence, the energy absorbed (in ergs/mole) is given as, (h = 6.62 × 10–27, ergs, c = 3 × 1010 cm s–1,, NA = 6.02 × 1023 mol–1), , 175, , 2., , Be2+ is isoelectronic with which of the following, ions?, (a) H+ +, (b) Li+ 2+, (c) Na, (d) Mg, (2014), , 3., , Isoelectronic– species, are, (a) CO, CN , NO+, C 2–, 2, , 4., , 5., , (b) CO–, CN, NO, C2–, (c) CO+, CN+, NO–, C2, (d) CO, CN, NO, C2, , (2000), , The ion that is isoelectronic with CO is, (a) CN –, (b) N 2+, (c) O2–, (d) N –2, , (1997), , Which one of the following is not isoelectronic with, O2–?, (a) Tl+, (b) Na+, (c) N3–, (d) F–, (1994), , 2.3, 6., , 7., , Developments Leading to the Bohr’s, Model of Atom, , Which of the following series of transitions in the, spectrum of hydrogen atom falls in visible region?, (a) Brackett series, (b) Lyman series, (c) Balmer series, (d) Paschen series, (NEET 2019), Calculate the energy in joule corresponding to light, of wavelength 45 nm., (Planck’s constant, h = 6.63 × 10–34 J s, speed of light,, c = 3 × 108 m s–1), (a) 6.67 × 1015, (b) 6.67 × 1011, –15, (c) 4.42 × 10, (d) 4.42 × 10–18, (2014), , www.neetujee.com, , 8, (a) 1.196 10, , 16, (c) 2.859 10, , , , , 2.859 105,  16, (d) 1.196 10, , (Karnataka NEET 2013), , (b), , 10. The energies E1 and E2 of two radiations are 25 eV, and 50 eV respectively. The relation between their, wavelengths i.e., 1 and 2 will be, (a) 1 = 2, (b) 1 = 22, 1, (2011), (c)  = 4, (d)   , 1, 2, 1, 2, 2, 11. The value of Planck’s constant is 6.63 × 10–34 J s., The velocity of light is 3.0 × 108 m s–1. Which value, is closest to the wavelength in nanometers of a, quantum of light with frequency of 8 × 1015 s–1 ?, (a) 2 × 10–25, (b) 5 × 10–18, 1, (c) 4 × 10, (d) 3 × 107, (2003), 12. For given energy, E = 3.03 × 10–19 joules, corresponding wavelength is, (h = 6.626 × 10–34 J sec, c = 3 × 108 m/sec), (a) 65.6 nm, (b) 6.56 nm, (c) 3.4 nm, (d) 656 nm, (2000), 13. What will be the longest wavelength line in Balmer, series of spectrum?, (a) 546 nm, (b) 656 nm, (c) 566 nm, (d) 556 nm, (1996), , www.mediit.in

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Structure of Atom, , 2.4, , Bohr’s Model for Hydrogen Atom, ,  Z2 , ,,  2 ,  n , certain conclusions are written. Which of them is, not correct?, (a) Equation can be used to calculate the change in, energy when the electron changes orbit., (b) For n = 1, the electron has a more negative, energy than it does for n = 6 which means, that the electron is more loosely bound in the, smallest allowed orbit., (c) The negative sign in equation simply means that, the energy of electron bound to the nucleus is, lower than it would be if the electrons were at, the infinite distance from the nucleus., (d) Larger the value of n, the larger is the orbit, radius., (NEET 2013), , 14. Based on equation E = – 2.178 × 10–18 J, , 15. According to the Bohr theory, which of the following, transitions in the hydrogen atom will give rise to the, least energetic photon?, (a) n = 6 to n = 1, (b) n = 5 to n = 4, (c) n = 6 to n = 5, (d) n = 5 to n = 3, (Mains 2011), 16. The energy of second Bohr orbit of the hydrogen, atom is –328 kJ mol–1; hence the energy of fourth, Bohr orbit would be, (a) – 41 kJ mol–1, (b) –82 kJ mol–1, (c) –164 kJ mol–1, (d) –1312 kJ mol–1 (2005), 17. The frequency of radiation emitted when the electron, falls from n = 4 to n = 1 in a hydrogen atom will be, (Given ionization energy of H = 2.18 × 10–18 J atom–1, and h = 6.626 × 10–34 J s), (a) 1.54 × 1015 s–1, (b) 1.03 × 1015 s–1, 15 –1, (c) 3.08 × 10 s, (d) 2.00 × 1015 s–1 (2004), , 9, , 21. In a Bohr’s model of an atom, when an electron, jumps from n = 1 to n = 3, how much energy will be, emitted or absorbed?, (a) 2.389 × 10–12 ergs (b) 0.239 × 10–10 ergs, (c) 2.15 × 10–11 ergs (d) 0.1936 × 10–10 ergs, (1996), 22. The radius of hydrogen atom in the ground state is, 0.53 Å. The radius of Li2+ ion (atomic number = 3), in a similar state is, (a) 0.53 Å, (b) 1.06 Å, (c) 0.17 Å, (d) 0.265 Å, (1995), 23. The energy of an electron in the nth Bohr orbit of, hydrogen atom is, 13.6, 13.6, eV, (a), (b), eV, (c), , n4, 13.6, n2, , eV, , n3, (d) 13.6, eV, n, , (1992), , 24. The spectrum of He is expected to be similar to that, (a) H, (b) Li+, (c) Na, (d) He+, (1988), 25. If r is the radius of the first orbit, the radius of nth, orbit of H-atom is given by, (a) rn2, (b) rn, (c) r/n, (d) r2n2, (1988), , 2.5, , Towards Quantum Mechanical Model of, the Atom, , 26. In hydrogen atom, the de Broglie wavelength, of an electron in the second Bohr orbit is, [Given that Bohr radius, a0 = 52.9 pm], (a) 211.6 pm, (b) 211.6  pm, (c) 52.9  pm, (d) 105.8 pm, (Odisha NEET 2019), , 18. In hydrogen atom, energy of first excited state is, –3.4 eV. Then find out K.E. of same orbit of hydrogen, atom., (a) +3.4 eV, (b) +6.8 eV, (c) –13.6 eV, (d) +13.6 eV, (2002), 19. Who modified Bohr’s theory by introducing, elliptical orbits for electron path?, (a) Rutherford, (b) Thomson, (c) Hund, (d) Sommerfeld (1999), , 27. A 0.66 kg ball is moving with a speed of 100 m/s., The associated wavelength will be, (h = 6.6 × 10–34 J s), (a) 6.6 × 10–32 m, (b) 6.6 × 10–34 m, –35, (c) 1.0 × 10 m, (d) 1.0 × 10–32 m, (Mains 2010), 28. If uncertainty in position and momentum are equal,, then uncertainty in velocity is, 1, 1, h, h (d), h (b) h, (c), (a), , 2, 2m , m , (2008), , 20. The Bohr orbit radius for the hydrogen atom, (n = 1) is approximately 0.530 Å. The radius for the, first excited state (n = 2) orbit is (in Å), (a) 4.77, (b) 1.06, (c) 0.13, (d) 2.12, (1998), , 29. The measurement of the electron position is, associated with an uncertainty in momentum,, which is equal to 1 × 10–18 g cm s–1. The uncertainty, in electron velocity is (mass of an electron is, 9 × 10–28 g), , www.neetujee.com, , www.mediit.in

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10, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (a) 1 × 105 cm s–1, (c) 1 × 109 cm s–1, , (b) 1 × 1011 cm s–1, (d) 1 × 106 cm s–1 (2008), , 30. Given : The mass of electron is 9.11 × 10–31 kg,, Planck constant is 6.626 × 10–34 J s, the uncertainty, involved in the measurement of velocity within a, distance of 0.1 Å is, (a) 5.79 × 105 m s–1, (b) 5.79 × 106 m s–1, 7, –1, (c) 5.79 × 10 m s, (d) 5.79 × 108 m s–1, (2006), 31. The uncertainty in momentum of an electron is, 1 × 10–5 kg m/s. The uncertainty in its position will be, (h = 6.62 × 10–34 kg m2/s), (a) 5.27 × 10–30 m, (b) 1.05 × 10–26 m, –28, (c) 1.05 × 10 m, (d) 5.25 × 10–28 m (1999), 32. The de Broglie wavelength of a particle with mass, 1 g and velocity 100 m/s is, (a) 6.63 × 10–35 m, (b) 6.63 × 10–34 m, –33, (c) 6.63 × 10 m, (d) 6.65 × 10–35 m (1999), 33. The position of both, an electron and a helium atom, is known within 1.0 nm. Further the momentum of, the electron is known within 5.0 × 10–26 kg m s–1., The minimum uncertainty in the measurement of, the momentum of the helium atom is, (a) 8.0 × 10–26 kg m s–1 (b) 80 kg m s–1, (c) 50 kg m s–1, (d) 5.0 × 10–26 kg m s–1, (1998), 34. Uncertainty in, position of an electron, (Mass = 9.1 × 10–28 g) moving with a velocity of, 3 × 104 cm/s accurate upto 0.001% will be, (Use h/(4) in uncertainty expression where, h = 6.626 × 10–27 erg second), (b) 7.68 cm, (a) 5.76 cm, (c) 1.93 cm, (d) 3.84 cm, (1995), 35. Which of the following statements do not form a, part of Bohr’s model of hydrogen atom?, (a) Energy of the electrons in the orbits are, quantized., (b) The electron in the orbit nearest the nucleus has, the lowest energy., (c) Electrons revolve in different orbits around the, nucleus., (d) The position and velocity of the electrons in the, orbit cannot be determined simultaneously., (1989), , 2.6 Quantum Mechanical Model of Atom, 36. 4d, 5p, 5f and 6p orbitals are arranged in the order of, decreasing energy. The correct option is, (a) 5f > 6p > 4d > 5p, (b) 5f > 6p > 5p > 4d, (c) 6p > 5f > 5p > 4d, (d) 6p > 5f > 4d > 5p, (NEET 2019), , www.neetujee.com, , 37. Orbital having 3 angular nodes and 3 total nodes is, (a) 5p, (b) 3d, (c) 4f, (d) 6d, (Odisha NEET 2019), 38. Which one is a wrong statement?, (a) Total orbital angular momentum of electron in, s-orbital is equal to zero., (b) An orbital is designated by three quantum, numbers while an electron in an atom is, designated by four quantum numbers., (c) The2 electronic configuration, 1 2p 1, 1 of N atom is, 2, , 1s, , 2s, , 2p, x, , y, , 2p, , z, , (d) The value of m for dz2 is zero., , (NEET 2018), , 39. Which one is the wrong statement?, , h, , 4, (b) Half filled and fully filled orbitals have greater, stability due to greater exchange energy, greater, symmetry and more balanced arrangement., (c) The energy of 2s-orbital is less than the energy, of 2p-orbital in case of hydrogen like atoms., h, (d) de-Broglie’s wavelength is given by   ,, mv, where m = mass of the particle, v = group, velocity of the particle., (NEET 2017), (a) The uncertainty principle is E  t , , 40. How many electrons can fit in the orbital for which, n = 3 and l = 1?, (a) 2, (b) 6, (c) 10, (d) 14, (NEET-II 2016), 41. Which of the following pairs of d-orbitals will have, electron density along the axes?, (a) dz2, dxz, (b) dxz, dyz, (c) d 2, d 2 2, (d) d , d 2 2, z, , x–y, , xy, , x –y, , (NEET-II 2016), 42. Two electrons occupying the same orbital are, distinguished by, (a) azimuthal quantum number, (b) spin quantum number, (c) principal quantum number, (d) magnetic quantum number., (NEET-I 2016), 43. Which is the correct order of increasing energy of, the listed orbitals in the atom of titanium?, (At. no. Z = 22), (a) 4s 3s 3p 3d, (b) 3s 3p 3d 4s, (c) 3s 3p 4s 3d, (d) 3s 4s 3p 3d, (2015), 2+, 44. The number of d-electrons in Fe (Z = 26) is not, equal to the number of electrons in which one of the, following?, (a) d-electrons in Fe (Z = 26), (b) p-electrons in Ne (Z = 10), , www.mediit.in

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Structure of Atom, (c) s-electrons in Mg (Z = 12), (d) p-electrons in Cl (Z = 17) (2015, Cancelled), 45. The angular momentum of electron in ‘d’ orbital is, equal to, (a) 2 3  (b) 0  (c) 6  (d) 2 , (2015, Cancelled), 46. What is the maximum number of orbitals that can, be identified with the following quantum numbers?, n = 3, l = 1, ml = 0, (a) 1, (b) 2, (c) 3, (d) 4, (2014), 47. What is the maximum numbers of electrons that, can be associated with the following set of quantum, numbers?, n = 3, l = 1 and m = –1, (a) 4, (b) 2, (c) 10, (d) 6, (NEET 2013), 48. The outer electronic configuration of Gd, (At. No. 64) is, (a) 4f 55d46s1, (b) 4f 75d16s2, 3, 5 2, (c) 4f 5d 6s, (d) 4f 45d56s1, (Karnataka NEET 2013), 49. Maximum number of electrons in a subshell with, l = 3 and n = 4 is, (a) 14, (b) 16, (c) 10, (d) 12, (2012), 50. The correct set of four quantum numbers for the, valence electron of rubidium atom (Z = 37) is, (a) 5, 1, 1, +1/2, (b) 6, 0, 0, +1/2, (c) 5, 0, 0, +1/2, (d) 5, 1, 0, +1/2, (2012), 51. The orbital angular momentum of a p-electron is, given as, (b) 3 h, h, (a), 2, 2 , h, h, (c) 3, (d) 6 (Mains 2012), 2, 2 , 52. The total number of atomic orbitals in fourth energy, level of an atom is, (a) 8, (b) 16, (c) 32, (d) 4, (2011), 53. If n = 6, the correct sequence for filling of electrons, will be, (a) ns  (n – 2)f  (n – 1)d  np, (b) ns  (n – 1)d  (n – 2)f  np, (c) ns  (n – 2)f  np  (n – 1)d, (d) ns  np  (n – 1)d  (n – 2)f, (2011), 54. Maximum number of electrons in a subshell of an, atom is determined by the following, , www.neetujee.com, , 11, , (a) 2l + 1, (c) 2n2, , (b) 4l – 2, (d) 4l + 2, , 55. Which of the following is not permissible, arrangement of electrons in an atom?, (a) n = 5, l = 3, m = 0, s = +1/2, (b) n = 3, l = 2, m = –3, s = –1/2, (c) n = 3, l = 2, m = –2, s = –1/2, (d) n = 4, l = 0, m = 0, s = –1/2, , (2009), , (2009), , 56. Consider the following sets of quantum numbers:, n l, m, s, (i) 3 0 0 +1/2, (ii) 2 2 1 +1/2, (iii) 4 3 –2, –1/2, (iv) 1 0 –1 –1/2, (v) 3 2 3 +1/2, Which of the following sets of quantum number is, not possible?, (a) (i), (ii), (iii) and (iv), (b) (ii), (iv) and (v), (c) (i) and (iii), (d) (ii), (iii) and (iv), (2007), 57. The orientation of an atomic orbital is governed by, (a) principal quantum number, (b) azimuthal quantum number, (c) spin quantum number, (d) magnetic quantum number., (2006), 58. The following quantum numbers are possible for, how many orbitals?, n = 3, l = 2, m = +2, (a) 1, (b) 2, (c) 3, (d) 4, (2001), 59. For which of the following sets of four quantum, numbers, an electron will have the highest energy?, n l m, s, (a) 3 2 1, +1/2, (b) 4 2 – 1 +1/2, (c) 4 1 0, –1/2, (d) 5 0 0, –1/2, (1994), 60. Electronic configuration of calcium atom can be, written as, (a) [Ne]4p2, (b) [Ar]4s2, 2, (c) [Ne]4s, (d) [Kr]4p2, (1992), 61. In a given atom no two electrons can have the same, values for all the four quantum numbers. This is, called, (a) Hund’s Rule, (b) Aufbau principle, (c) Uncertainty principle, (d) Pauli’s Exclusion principle., (1991), , www.mediit.in

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12, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 62. For azimuthal quantum number l = 3, the maximum, number of electrons will be, (a) 2, (b) 6, (c) 0, (d) 14, (1991), 63. The order of filling of electrons in the orbitals of an, atom will be, (a) 3d, 4s, 4p, 4d, 5s, (b) 4s, 3d, 4p, 5s, 4d, (c) 5s, 4p, 3d, 4d, 5s, (d) 3d, 4p, 4s, 4d, 5s, (1991), , quantum number 2 and azimuthal quantum number, 1 are, (a) 2, (b) 4, (c) 6, (d) 8, (1990), , 64. The electronic configuration of Cu (atomic number, 29) is, (a) 1s2 2s22p6 3s23p6 4s23d9, (b) 1s2 2s22p6 3s23p63d10 4s1, (c) 1s2 2s22p6 3s23p6 4s24p6 5s25p1, (d) 1s2 2s22p6 3s23p6 4s24p63d3, (1991), 65. The total number of electrons that can be, accommodated in all the orbitals having principal, , 66. An ion has 18 electrons in the outermost shell, it is, (a) Cu+, (b) Th4+, (c) Cs+, (d) K+, (1990), 67. Number of unpaired electrons in N2+ is/are, (a) 2, (b) 0, (c) 1, (d) 3, (1989), 68. The maximum number of electrons in a subshell is, given by the expression, (a) 4l – 2, (b) 4l + 2, (c) 2l + 2, (d) 2n2, (1989), 69. The number of spherical nodes in 3p orbitals are/is, (a) one, (b) three, (c) none, (d) two, (1988), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., 61., , (a), (c), (d), (a), (c), (a), (d), , 2., 12., 22., 32., 42., 52., 62., , (b), (d), (c), (c), (b), (b), (d), , 3., 13., 23., 33., 43., 53., 63., , (a), (b), (c), (d), (c), (a), (b), , 4., 14., 24., 34., 44., 54., 64., , (a), (b), (b), (c), (d), (d), (b), , 5., 15., 25., 35., 45., 55., 65., , (a), (c), (a), (d), (c), (b), (c), , 6., 16., 26., 36., 46., 56., 66., , (c), (b), (b), (b), (a), (b), (a), , 7., 17., 27., 37., 47., 57., 67., , (d), (c), (c), (c), (b), (d), (c), , 8., 18., 28., 38., 48., 58., 68., , (a), (a), (c), (c), (b), (a), (b), , 9., 19., 29., 39., 49., 59., 69., , (a), (d), (c), (c), (a), (b), (a), , 10., 20., 30., 40., 50., 60., , (b), (d), (b), (a), (c), (b), , Hints & Explanations, 1., , (a) :, , 175, , Lu, Number of protons = Number of, electrons = Atomic number = 71, Number of neutrons = Mass number – Atomic number, = 175 – 71 = 104, 2., , 71, , (b) : Species, No. of electrons, 2+, Be, 2, H+, 0, Li+, 2, Na+, 10, Mg2+, 10, 3. (a) : Species having same no. of electrons are called, isoelectronic species., –, +, 2–, The no. of electrons in CO = CN = NO = C2 = 14. So, these, are isoelectronic species., 4. (a) : Since both CO and CN– have 14 electrons,, therefore these are isoelectronic species (i.e. having same, number of electrons)., , www.neetujee.com, , 5. (a) : The number of electrons in O2–, N3–, F– and, Na+ is 10 each, but number of electrons in Tl + is 80., 6. (c) : Lyman series : UV region, Balmer series : Visible region, Paschen series : IR region, Brackett series : IR region, hc, 7. (d) : E  [Given,  = 45 nm = 45 × 10–9 m], , On putting the given values in the equation, we get, 6.63 1034  3 108, E,  4.42 1018 J, 45 109, 8., , (a) : c = , c 3 1017,  ,  50 nm, , , 6 1015, , www.mediit.in

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Structure of Atom, 9., , 13, , (a) : We know that, E , 27, , , , hcNA, , , 10, , 17. (c) : E = h or  = E/h 19, 21.8 10, For H atom, E , J atom1, , 23, , 6.62 10, ,  3 10  6.02 10, , 1.1955 10, 1.196 108, 8, , , , , , , hc, hc, 10. (b) : E , and, E, , , ;, 1, , 1, , ergs mol, , , 2, , 1, , E, , 1, , c 3 10  37.5 10 m,  ,  8 1015, = 37.5 nm  4 × 101 nm, hc, 6.626 1034  3 108, 12. (d) : E , , , 3.03 1019, = 656 nm, 13. (b) : The longest wavelength means the lowest, energy., 1 We know that relation for wavelength,  R  1 1 , , H 2, , n2 , n, 2, ,  e2 , , 2, , 18. (a) : Kinetic energy  mv    ,  2m, nh, 2, , 2e2 , v, , , , , , nh , 2, 2, 4, , , 2 me, e 2, Total energy, En = ,      2m = –K. E ., n2h2, , 9, , 8, , 2, , , , 2, , 11. (c) : Applying  = c/,, , –1, , (RH, Rydberg constant = 109677 cm ), For n1 = 2, n2 = 3, 1, 1, 1 ,  109677, ,  15233, 2, 2, , 2 3 , 1, or,  , = 6.56 × 10–5 cm, 15233, = 6.56 × 10–7 m = 656 nm, 14. (b) : The electron is more tightly bound in the, smallest allowed orbit., 15. (c) : We know that, 1 1 , E     , where n2  n1,  n12, , 6.626 1034, ,  2, , 4, 12 , 15 1,  3.08 10 s, 1, ,   2, , 1, , 1, , 20.44 1019, , n2, 1 1,   = 20.44 × 10–19 J atom–1, , 2, , 1 hc 2 2, E2  1 hc 1, 25 2, 1 2, or, , or,  , 50 1, 2 , , n22 , ,  n = 6 to n = 5 will give least energetic photon., 2, Z, 16. (b) : En  K   , n , Z = 1 for hydrogen; n = 2, = –328 kJ mol–1; K = 4 × 328, K  1, E2, , E, 2, 4, K  1, 1, E,  E = – 4 × 328 ×, = – 82 kJ mol–1, 4, 4, 16, 16, , www.neetujee.com, , E  21.8  10, , 19 , ,  nh , , , , Kinetic energy = –En, Energy of first excited state is –3.4 eV.,  Kinetic energy of same orbit (n = 2) will be +3.4 eV., 19. (d) : Sommerfeld modified Bohr’s theory, considering that in addition to circular orbits electrons, also move in elliptical orbits., 20. (d) : For nth orbit of ‘H’ atom, rn = n2 × r1,  radius of 2nd Bohr’s orbit., r2 = 4 × r1 = 4 × 0.53 = 2.12 Å, 21. (d)1312, : Energy of an atom when n = 1, = – 1312 kJ mol–1, E =, 2, 1, (1), 1312, Similarly energy when n = 3, (E3) = , , 32, = – 145.7 kJ mol–1, The energy absorbed when an electron jumps from, n = 1 to n = 3,, E3 – E1 = – 145.7 – (– 1312) = 1166.3 kJ mol–1, 1166.3, =, = 193.6 × 10–23 kJ, 23, 6.023  10, = 193.6 × 10–20 J, [1 joule = 107 ergs], –13, –10,  193.6 × 10 ergs = 0.1936 × 10 ergs, 22. (c) : Due to ground state, state of hydrogen atom, (n) = 1; Radius of hydrogen atom (r) = 0.53 Å, Atomic no. of Li (Z) = 3, n2, (1)2, Now, radius of Li2+ ion  r  Z  0.53  3  0.17Å, 23. (c) : Energy of an electron in nth Bohr orbit of, 13.6, hydrogen atom  n2 eV., 24. (b) : Both He and Li+ contain 2 electrons each., 25. (a) : Radius of nth orbit of H-atom = r0n2, where r0 = radius of the first orbit., , www.mediit.in

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14, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 26. (b) : Bohr radius, a0 = 52.9 pm, n = 2, rn = n2a0 = (2)2a0 = 4 × 52.9 pm = 211.6 pm, The angular momentum of an electron in a given, stationary state can be expressed as in equation,, h h, h, mvr = n . = 2  ,  mvr  h, ... (i), 2 , 2, de-Broglie equation,, h, , ; mv = h, ... (ii), mv, From equations (i) and (ii), we get  = r, Putting the value of r,  = 211.6  pm, h, 27. (c) : According to de-Broglie equation,  , mv, Given, h = 6.6 × 10–34 J s ; m = 0.66 kg ; v = 100 m s–1, 6.6 1034, 35,  ,  1  10 m, 0.66 100, 28. (c) : From Heisenberg uncertainty principle,, h, h, p x , or mv  x , 4, 4, h, 2, or (mv) , ( x  p), 4, 1, h, or v , 2m , 29. (c) : Uncertainty in momentum (mv), = 1 × 10–18 g cm s–1, Uncertainty in velocity (v), 1  1018, 9, 1,  1.1  10 cm s, , 9  1028, 30. (b) : x·mv = h/4, 6.626 1034, 0.1 1010  9.11 1031  v , 4  3.143, , h, , 4, As, x =1.0 nm for both electron and helium atom, so p, is also same for both the particles., Thus, uncertainty in momentum of the helium atom is, also 5.0 × 10–26 kg m s–1., 34. (c) : Mass of an electron (m) = 9.1 × 10–28 g, Velocity of electron (v) = 3 × 104 cm/s, 0.001, Accuracy = 0.001% =, and, 100, Planck’s constant (h) = 6.626 × 10–27 erg-second., We know that actual velocity of the electron, 0.001, (v) = 3 × 104 ×, = 0.3 cm/s, 100, Therefore, uncertainty in the position of the electron,, h, 6.626 10 27, (x)  , = 1.93 cm, 4mv 4  (9.1 1028 )  0.3, 35. (d) : It is Heisenberg’s uncertainty principle and not, Bohr’s postulate., 36. (b) : Higher the value of (n + l) for an orbital, higher, is its energy. However, if two different types of orbitals, have same value of (n + l), the orbital with lower value of, n has lower energy. Therefore, decreasing order of energy, of the given orbitals is 5f > 6p > 5p > 4d., 37. (c) : Number of spherical/radial nodes in any, orbital = n – l – 1, Number of planar/angular nodes in orbital = l = 3,  Total number of nodes in any orbital = n –1 = 3,  n=4, Thus, the orbital is 4f., 38. (c) : According to Hund’s rule of maximum, multiplicity, the correct configuration of ‘N’ is, , 6.626 1034, ,  v , , 0.1 1010  9.11 1031  4  3.143, = 5.79 × 106 m s–1, h, 31. (a) : x × p =, 4, (Heisenberg uncertainty principle),  x =, , i.e., x  p , , 6.62 1034, –30, 4  3.14 105 = 5.27 × 10 m, , h, 32. (c) :  , mv, , 6.63 1027 erg sec, , 1g 104 cm/s, = 6.63 × 10–31 cm = 6.63 × 10–33 m, 33. (d) : According to uncertainty principle the product, of uncertainty in position and uncertainty in momentum, is constant for a particle., , www.neetujee.com, , 39. (c) : In case of hydrogen like atoms, energy depends, on the principal quantum number only. Hence, 2s-orbital, will have energy equal to 2p-orbital., 40. (a) : For n = 3 and l = 1, the subshell is 3p and a, particular 3p orbital can accommodate only 2 electrons., 41. (c) : dx2 – y2 and dz2 orbitals have electron density, along the axes while dxy, dyz and dxz orbitals have electron, density in between the axes., 42. (b) : For the two, , electrons occupying the same, orbital1values of, n l and ml are same but ms is different,, i.e.,  and  1 ., 2, 2, 43. (c) : Ti(22) : 1s22s22p63s23p64s23d2,  Order of increasing energy is 3s, 3p, 4s, 3d., , www.mediit.in

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Structure of Atom, , 15, , 44. (d) : Number of d-electrons in Fe2+ = 6, Number of p-electrons in Cl = 11, 45. (c) : Angular momentum  l(l  1) , For d-orbital, l = 2, Angular momentum  2(2  1)  , , 6, 46. (a) : Only one orbital, 3pz has following set of, quantum numbers, n = 3, l = 1 and ml = 0., 47. (b) : The orbital associated with n = 3, l = 1 is 3p. One, orbital (with m = –1) of 3p-subshell can accommodate, maximum 2 electrons., 48. (b) : The electronic configuration of 64Gd is, [Xe]4f 75d16s2., 49. (a) : l = 3 and n = 4 represents 4f. So, total number, of electrons in a subshell = 2(2l + 1) = 2(2 × 3 + 1) = 14, electrons. Hence, f-subshell can contain maximum 14, electrons., 50. (c) : Rb(37) : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1, For 5s, n = 5, l = 0, m = 0, s = +1/2 or –1/2, 51. (a) : Orbital angular momentum (m), h, 2, For p-electrons; l = 1,  l(l 1), , Thus, m  1(11), , h, 2, , , , 2h h, , 2 2 , , (iv) is not possible as value of m varies from – l ... +l., (v) is not possible as value of m varies from, –l ... +l, it can never be greater than l., 57. (d) : Principal quantum number represents the, name, size and energy of the shell to which the electron, belongs. Azimuthal quantum number describes the, spatial distribution of electron cloud and angular, momentum. Magnetic quantum number describes, the orientation or distribution of electron cloud. Spin, quantum number represents the direction of electron, spin around its own axis., 58. (a) : n = 3, l = 2, m = +2, It symbolises one of the five d-orbitals (3d)., m = +2 +1, , 0, , –1 –2, , 59. (b) : Energy of electron depends on the value of, (n + l). The subshell are 3d, 4d, 4p and 5s, out of which 4d, has highest energy., 60. (b) : Atomic no. of Ca = 20, 2,  Electronic configuration of Ca = [Ar]4s, 61. (d) : This is a Pauli’s exclusion principle., 62. (d) : l = 3 means f-subshell, Maximum no. of electrons in f-subshell = 14, f-subshell =, , 52. (b) : Total number of atomic orbitals in any energy, level is given by n2., 53. (a), 54. (d) : For a given shell, l,, the number of subshells, ml = (2l + 1), Since each subshell can accommodate 2 electrons of, opposite spin, so maximum number of electrons in a, subshell = 2(2l + 1) = 4l + 2., 55. (b) : In an atom, for any value of n, the values of, l = 0 to (n – 1)., For a given value of l, the values of ml = –l to 0 to +l and, the value of s = +1/2 or –1/2., In option (b), l = 2 and ml = –3, This is not possible, as values of ml which are possible for, l = 2 are –2, –1, 0, +1 and +2 only., 56. (b) : (i) represents an electron in 3s orbital., (ii) is not possible as value of l varies from 0, 1, ... (n – 1)., (iii) represents an electron in 4f orbital., , 63. (b) : Higher the value of (n + l) for an orbital, higher, is its energy. However, if two different types of orbitals, have same value of (n + l), the orbital with lower value, of n has lower energy., 64. (b) : Electronic configuration of Cu is, 1s22s22p63s23p63d104s1., 65. (c) : n = 2, l = 1, It means 2p-orbitals., Total no. of electrons that can be accommodated in all, the 2p orbitals = 6, 66. (a) : Cu+ ion has 18 electrons in its outermost shell., Electronic configuration of Cu+ is 1s22s22p6 3s23p63d10., 67. (c) : N2+ = 1s22s22p1,  No. of unpaired electrons = 1, 68. (b) : No. of orbitals in a subshell = 2l + 1,  No. of electrons = 2(2l + 1) = 4l + 2, 69. (a) : No. of radial nodes in 3p-orbital = n – l – 1, =3–1–1=1, , , , www.neetujee.com, , www.mediit.in

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, , , , , , CHAPTER, , 3, , 3.4, 1., , Classification of, Elements and Periodicity, in Properties, , Nomenclature of Elements with Atomic, Numbers > 100, , Identify the incorrect match., Name, IUPAC Official Name, (A) Unnilunium, (i) Mendelevium, (B) Unniltrium, (ii) Lawrencium, (C) Unnilhexium, (iii) Seaborgium, (D) Unununnium, (iv) Darmstadtium, (a) (A), (i), (b) (B), (ii), (c) (C), (iii), (d) (D), (iv), (NEET 2020), , 3.5, , Electronic Configurations of Elements, and The Periodic Table, , 2., , The element Z = 114 has been discovered recently. It, will belong to which of the following family/group, and electronic configuration?, (a) Carbon family, [Rn] 5f 14 6d10 7s2 7p2, (b) Oxygen family, [Rn] 5f 14 6d10 7s2 7p4, (c) Nitrogen family, [Rn] 5f14 6d10 7s2 7p6, (d) Halogen family, [Rn] 5f 14 6d10 7s2 7p5, (NEET 2017), , 3., , An atom has electronic configuration, 1s2 2s2 2p6 3s2 3p6 3d3 4s2, you will place it in, (a) fifth group, (b) fifteenth group, (c) second group, (d) third group., (2002), , 4., , 5., , The electronic configuration of an element is, 1s2 2s2 2p6 3s2 3p3. What is the atomic number of the, element, which is just below the above element in, the periodic table?, (a) 36, (b) 49, (c) 33, (d) 34, (1995), If the atomic number of an element is 33, it will be, placed in the periodic table in the, (a) first group, (b) third group, (c) fifth group, (d) seventh group. (1993), , www.neetujee.com, , 6., , The electronic configuration of four elements are, given below. Which elements does not belong to the, same family as others?, (a) [Xe]4f 145d104s2, (c) [Ne]3s23p5, , 3.7, , (b) [Kr]4d105s2, (d) [Ar]3d104s2, , (1989), , Periodic Trends in Properties of Elements, , 7., , For the second period elements the correct increasing, order of first ionization enthalpy is, (a) Li < Be < B < C < O < N < F < Ne, (b) Li < Be < B < C < N < O < F < Ne, (c) Li < B < Be < C < O < N < F < Ne, (d) Li < B < Be < C < N < O < F < Ne (NEET 2019), , 8., , Match the oxide given in column I with its property, given in column II., Column I, Column II, (i) Na2O, A. Neutral, (ii) Al2O3, B. Basic, (iii) N2O, C. Acidic, (iv) Cl2O7, D. Amphoteric, Which of the following options has all correct pairs?, (a) (i)-B, (ii)-A, (iii)-D, (iv)-C, (b) (i)-C, (ii)-B, (iii)-A, (iv)-D, (c) (i)-A, (ii)-D, (iii)-B, (iv)-C, (d) (i)-B, (ii)-D, (iii)-A, (iv)-C (Odisha NEET 2019), , 9., , Which of the following oxides is most acidic in nature?, (a) MgO, (b) BeO, (c) BaO, (d) CaO, (NEET 2018), , 10. In which of the following options the order of, arrangement does not agree with the variation of, property indicated against it?, (a) I < Br < Cl < F (increasing electron gain, enthalpy), (b) Li < Na < K < Rb (increasing metallic radius), (c) Al3+ < Mg2+ < Na+ < F– (increasing ionic size), (d) B < C < N < O (increasing first ionisation, enthalpy), (NEET-I 2016), , www.mediit.in

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17, , Classification of Elements and Periodicity in Properties, 11. The formation of the oxide ion, O 2–, (g) from oxygen, atom requires first an exothermic and then an, endothermic step as shown below :, O( g )  e   O(g );  f H  141 kJ mol 1, , 16. What is the value of electron gain enthalpy of Na+ if, IE1 of Na = 5.1 eV?, (a) –5.1 eV, (b) –10.2 eV, (c) +2.55 eV, (d) +10.2 eV(Mains 2011), , O(g) e   O2( g ;)  f H  780 kJ mol 1, , 17. Which of the following oxides is amphoteric?, (a) SnO2, (b) CaO, , , , Thus, process of formation of O2– in gas phase is, unfavourable even though O2– is isoelectronic with, neon. It is due to the fact that,, (a) O– ion has comparatively smaller size than, oxygen atom, (b) oxygen is more electronegative, (c) addition of electron in oxygen results in larger, size of the ion, (d) electron repulsion outweighs the stability gained, by achieving noble gas configuration. (2015), 12. Which of the following orders of ionic radii is, correctly represented?, (a) H– > H+ > H, (b) Na+ > F– > O2–, (c) F– > O2– > Na+, (d) Al3+ > Mg2+ > N3–, (2014), 13. Which one of the following arrangements represents, the correct order of least negative to most negative, electron gain enthalpy for C, Ca, Al, F and O?, (a) Al < Ca < O < C < F, (b) Al < O < C < Ca < F, (c) C < F < O < Al < Ca, (d) Ca < Al < C < O < F (Karnataka NEET 2013), 14. In which of the following arrangements the given, sequence is not strictly according to the property, indicated against it?, (a) HF < HCl < HBr < HI : increasing acidic strength, (b) H O < H S < H Se < H Te : increasing pK values, 2, , 2, , 2, , 2, , a, , (c) NH3 < PH3 < AsH3 < SbH3 : increasing acidic, character, (d) CO2 < SiO2 < SnO 2 < PbO2 : increasing oxidising, power, (Mains 2012), 15. Identify the wrong statement in the following., (a) Amongst isoelectronic species, smaller the, positive charge on the cation, smaller is the, ionic radius., (b) Amongst isoelectronic species, greater the, negative charge on the anion, larger is the ionic, radius., (c) Atomic radius of the elements increases as one, moves down the first group of the periodic table., (d) Atomic radius of the elements decreases as one, moves across from left to right in the 2nd period, of the periodic table., (2012), , www.neetujee.com, , (c) SiO2, , (d) CO2, , (Mains 2011), , 18. The correct order of the decreasing ionic radii, among the following isoelectronic species is, (a) Ca2+ > K+ > S2– > Cl–, (b) Cl– > S2– > Ca2+ > K+, (c) S2– > Cl– > K+ > Ca2+, (d) K+ > Ca2+ > Cl– > S2–, (2010), 19. Which of the following represents the correct order, of increasing electron gain enthalpy with negative, sign for the elements O, S, F and Cl?, (a) Cl < F < O < S, (b) O < S < F < Cl, (c) F < S < O < Cl, (d) S < O < Cl < F, (2010, 2005), 20. Among the elements Ca, Mg, P and Cl, the order of, increasing atomic radii is, (a) Mg < Ca < Cl < P (b) Cl < P < Mg < Ca, (c) P < Cl < Ca < Mg (d) Ca < Mg < P < Cl, (Mains 2010), 21. Among the following which one has the highest, cation to anion size ratio?, (a) CsI, (b) CsF, (c) LiF, (d) NaF (Mains 2010), 22. Amongst the elements with following electronic, configurations, which one of them may have the, highest ionisation energy?, (a) Ne [3s2 3p2], (b) Ar [3d10 4s2 4p3], (c) Ne [3s2 3p1], (d) Ne [3s2 3p3], (2009), 23. Identify the correct order of the size of the following., (a) Ca2+ < K+ < Ar < Cl– < S2–, (b) Ar < Ca2+ < K+ < Cl– < S2–, (c) Ca2+ < Ar < K+ < Cl– < S2–, (d) Ca2+ < K+ < Ar < S2– < Cl–, , (2007), , 24. With which of the following electronic configuration, an atom has the lowest ionisation enthalpy?, (a) 1s2 2s2 2p3, (b) 1s2 2s2 2p5 3s1, (c) 1s2 2s2 2p6, (d) 1s2 2s2 2p5, (2007), 25. Which one of the following ionic species has the, greatest proton affinity to form stable compound?, (a) NH 2–, (b) F–, –, (c) I, (d) HS–, (2007), , www.mediit.in

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18, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 26. Which of the following is the most basic oxide?, (a) SeO2, (b) Al2O3, (c) Sb2O3, (d) Bi2O3, (2006), 27. What is the correct relationship between the pH of, isomolar solutions of sodium oxide, Na2O (pH1),, sodium sulphide, Na S2 (pH ),2 sodium selenide,, Na2Se (pH3) and sodium telluride Na2Te (pH4)?, (b) pH1 > pH2  pH3 > pH4, (c) pH1 < pH2 < pH3 < pH4, (2005), , 28. Ionic radii are, (a) inversely proportional to effective nuclear charge, (b) inversely proportional to square of effective, nuclear charge, (c) directly proportional to effective nuclear charge, (d) directly proportional to square of effective, nuclear charge., (2004), 29. The ions O2–, F–, Na+, Mg2+ and Al3+ are isoelectronic., Their ionic radii show, (a) a significant increase from O2– to Al3+, (b) a significant decrease from O2– to Al3+, (c) an increase from O2– to F– and then decrease, from Na+ to Al3+, (d) a decrease from O2– to F– and then increase, (2003), from Na+ to Al3+., 30. Which of the following order is wrong?, (a) NH3 < PH3 < AsH3 – acidic, (b) Li < Be < B < C – 1st IP, (c) Al2O3 < MgO < Na2O < K2O – basic, (d) Li+ < Na+ < K+ < Cs+ – ionic radius., , 35. Which of the following ions is the largest in size?, (a) K+, (b) Ca2+, (c) Cl–, (d) S2–, (1996), 36. Which of the following has the smallest size?, (a) Al3+, (b) F–, +, (c) Na, (d) Mg2+, (1996), , (a) pH1 > pH2 > pH3 > pH4, , (d) pH1 < pH2 < pH3  pH4, , 34. Which one of the following is correct order of the, size of iodine species?, (a) I+ > I– > I, (b) I– > I > I+, –, +, (c) I > I > I, (d) I > I+ > I–, (1997), , (2002), , 31. Correct order of 1st ionisation potential among, following elements Be, B, C, N, O is, (a) B < Be < C < O < N, (b) B < Be < C < N < O, (c) Be < B < C < N < O, (d) Be < B < C < O < N, (2001), , 37. Among the following oxides, the one which is most, basic is, (a) ZnO, (b) MgO, (c) Al2O3, (d) N2O5, (1994), 38. Which of the following has largest size?, (a) Na, (b) Na+, –, (c) Na, (d) Can’t be predicted., (1993), 39. Na+, Mg2+, Al3+ and Si4+ are isoelectronic. The order, of their ionic size is, (a) Na+ > Mg2+ < Al3+ < Si4+, (b) Na+ < Mg2+ > Al3+ > Si4+, (c) Na+ > Mg2+ > Al3+ > Si4+, (d) Na+ < Mg2+ > Al3+ < Si4+, (1993), 40. In the periodic table from left to right in a period,, the atomic volume, (a) decreases, (b) increases, (c) remains same, (d) first decreases then increases., (1993), 41. Which electronic configuration of an element has, abnormally high difference between second and, third ionization energy?, (a) 1s2, 2s2, 2p6, 3s1, (b) 1s2, 2s2, 2p6, 3s1, 3p1, (c) 1s2, 2s2, 2p6, 3s2, 3p2, (d) 1s2, 2s2, 2p6, 3s2, (1993), , 32. Which of the following elements has the maximum, electron affinity?, (a) I, (b) Br, (c) Cl, (d) F, (1999), , 42. One of the characteristic properties of non-metals is, that they, (a) are reducing agents, (b) form basic oxides, (c) form cations by electron gain, (d) are electronegative., (1993), , 33. The first ionization potentials (eV) of Be and B, respectively are, (a) 8.29, 8.29, (b) 9.32, 9.32, (c) 8.29, 9.32, (d) 9.32, 8.29, (1998), , 43. Which one of the following has minimum value of, cation/anion ratio?, (a) NaCl, (b) KCl, (c) MgCl2, (d) CaF2, (1993), , www.neetujee.com, , www.mediit.in

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Classification of Elements and Periodicity in Properties, , 19, , 44. Which of the following sets has strongest tendency, to form anions?, (a) Ga, Ni, Tl, (b) Na, Mg, Al, (c) N, O, F, (d) V, Cr, Mn, (1993), , 46. In the periodic table, with the increase in atomic, number, the metallic character of an element, (a) decreases in a period and increases in a group, (b) increases in a period and decreases in a group, (c) increases both in a period and the group, (d) decreases in a period and the group., (1989), , 45. Elements of which of the following groups will form, anions most readily?, (a) Oxygen family, (b) Nitrogen family, (c) Halogens, (d) Alkali metals (1992), , 47. Which of the following atoms will have the smallest, size?, (a) Mg, (b) Na, (c) Be, (d) Li, (1989), , ANSWER KEY, , 1., 11., 20., 30., 40., , (d), (d), (b), (b), (d), , 2., 12., 21., 31., 41., , (a) 3., (None), (b) 22., (a) 32., (d) 42., , (a), (d), (c), (d), , 4., 13., 23., 33., 43., , (c), (d), (a), (d), (c), , 5., 14., 24., 34., 44., , (c), (b), (b), (b), (c), , 6., 15., 25., 35., 45., , (c), (a), (a), (d), (c), , 7., 16., 26., 36., 46., , (c), (a), (d), (a), (a), , 8., 17., 27., 37., 47., , (d), (a), (a), (b), (c), , 9., 18., 28., 38., , (b), (c), (a), (c), , 10., 19., 29., 39., , (a,d), (b), (b), (c), , Hints & Explanations, 1. (d) : Unnilunium – Mendelevium  (a)-(i), Unniltrium – Lawrencium  (b)-(ii), Unnilhexium – Seaborgium  (c)-(iii), Unununnium – Roentgenium  (d) (iv), , and decrease in atomic radii. However, abnormal values, are observed for Be, N and Ne due to extra stability of, half filled and fully filled orbitals. Thus, the actual order, is, Li < B < Be < C < O < N < F < Ne., , 2. (a) : The electronic configuration of the element, with Z = 114 (Flerovium) is [Rn]5f14 6d107s27p2., Hence, it belongs to carbon family which has the same, outer electronic configuration., , 8. (d) : Na2O - Basic oxide, Al2O3 - Amphoteric oxide,, N2O - Neutral oxide, Cl2O7 - Acidic oxide., , 3., , (a) : The electronic configuration of an atom:, 1s2 2s2 2p6 3s2 3p6 3d3 4s2, In the configuration, the last electron of the atom is, filled in d-subshell as 3d3. Thus, this element belongs to, d-block of the periodic table with group no. VB or 5., 4. (c) : Atomic number of the given element is 15 and, it belongs to group 15. Therefore atomic number of the, element below the above element = 15 + 18 = 33., 5. (c) : Electronic configuration of an element with, Z = 33 is 1s22s22p63s23p63d104s24p3., Hence, it lies in VA or 15th group., 6. (c) : Elements (a), (b) and (d) belong to the same, group since each one of them has two electrons in valence, shell. In contrast, element (c) has seven electrons in the, valence shell, and hence it lies in other group., 7. (c) : As we move across a period, ionisation, enthalpy increases, because of increased nuclear charge, , www.neetujee.com, , 9. (b) : In metals, on moving down the group, metallic, character increases, so basic nature increases hence most, acidic will be BeO., 10. (a, d) : The correct order of increasing negative, electron gain enthalpy is : I < Br < F < Cl due to, electron-electron repulsion in small sized F atom and, the correct order of increasing first ionisation enthalpy is, B < C < O < N due to extra stability of half-filled orbitals, in N-atom., 11. (d), 12. (None) : Cations lose electrons and are smaller in, size than the parent atom, whereas anions add electrons, and are larger in size than the parent atom. Hence, the order, is H– > H > H+., For isoelectronic species, the ionic radii decreases with, increase in atomic number i.e., nuclear charge. Hence,, the correct orders are, O2– > F– > Na+ and N3– > Mg2+ > Al3+., , www.mediit.in

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20, , 13. (d) Electron gain enthalpy becomes less negative, from top to bottom in a group while it becomes more, negative from left to right within a period., 14. (b) : Acidic strength of hydrides increase with, increase in molecular mass., Thus, order of acidic strength is, HF < HCl < HBr < HI, H2O < H2S < H2Se < H2Te, NH3 < PH3 < AsH3 < SbH3, and as acidic strength increases, pKa decreases. Thus, order of pKa, H2O > H2S > H2Se > H2Te, 15. (a) : As positive charge on the cation increases,, effective nuclear charge increases. Thus, atomic size, decreases., 16. (a) : Na  Na+ + e–; H = 5.1 eV, Na+ + e–  Na ; H = –5.1 eV, 17. (a) : SnO2 reacts with acid as well as base. So, SnO2, is an amphoteric oxide., SnO2 + 4HCl  SnCl2 + 2H2O, SnO2 + 2NaOH  Na2SnO3 + H2O, CaO is basic in nature while SiO2 and CO2 are acidic in, nature., 18. (c) : S2– > Cl– > K+ > Ca2+, Among isoelectronic species, ionic radii increases, with increase in negative charge. This happens because, effective nuclear charge (Zeff) decreases., Similarly, ionic radii decreases with increase in positive, charge as Zeff increases., 19. (b) : Cl atom has the highest electron affinity in the, periodic table. F being a member of group 17 has higher, electron gain enthalpy than S which belongs to group 16., This in turn is higher than the electron affinity of O atom., Thus, Cl > F > S > O, It is worth noting that the electron gain enthalpy of, oxygen and fluorine, the members of the second period,, have less negative values of electron gain enthalpy than, the corresponding elements sulphur and chlorine of the, third period., This is due to small size of the atoms of oxygen and, fluorine. As a result, there is a strong inter-electronic, repulsion when extra electron is added to these atoms,, i.e., electron density is high and the addition of an extra, electron is not easy., 20. (b) : The atomic radii decrease on moving from left, to right in a period, thus order of sizes for Cl, P and Mg, is Cl < P < Mg. Down the group size increases. Thus,, overall order is Cl < P < Mg < Ca., 21. (b) : The cation to anion size ratio will be maximum, when the cation is of largest size and the anion is of, , www.neetujee.com, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , smallest size. Among the given species, Cs+ has maximum, size among given cations and F– has smallest size among, given anions, thus CsF has highest rc/ra ratio., 22. (d) : Among options (a), (c) and (d), option (d) has, the highest ionisation energy because of extra stability, associated with half-filled 3p-orbital. In option (b), the, presence of 3d10 electrons offers shielding effect, as a result, the 4p3 electrons do not experience much nuclear charge, and hence, the electrons can be removed easily., 23. (a) : Among isoelectronic ions, ionic radii of anions, is more than that of cations. Further size of the anion, increases with increase in negative charge and size of the, cation decreases with increase in positive charge., 24. (b) : The larger the atomic size, smaller is the value, of the ionisation enthalpy. Again higher the screening, effect, lesser is the value of ionisation potential. Hence,, option (b) has lowest ionisation enthalpy., 25. (a) : In going from left to right across a period in the, periodic table, the basicity (i.e., proton affinity) decreases, as the electronegativity of the atom possessing the lone, pair of electrons, increases. Hence, basicity of NH–2 is, –, higher than F . On moving down a group, as the atomic, size increases, basicity decreases. Hence, F– is more basic, than I– and HO– is more basic than HS–. Hence, among, the given ionic species, NH 2 – has maximum proton, affinity., 26. (d) : SeO2  acidic oxide,, Al2O3, Sb2O3  amphoteric,, Bi2O3  basic oxide., 27. (a) : Na2O, Basic character, Na2S, decreases down the group, Na2Se, Na2Te, pH  basic character, Hence, pH 1 > pH2 > pH3 > pH4, 28. (a), 29. (b) : Amongst isoelectronic ions, ionic radii of, anions is more than that of cations. Further size of the, anion increases with increase in –ve charge and size of, cation decreases with increase in +ve charge. Hence,, correct order is O2– > F– > Na+ > Mg2+ > Al3+., 30. (b) : Li, Be, B, C - these elements belong to the same, period. Generally the value of 1st ionisation potential, increases on moving from left to right in a period, since, the nuclear charge of the elements also increase in the, same direction. But the ionisation potential of boron, (B  2s2 2p1) is lower than that of beryllium (Be  2s2),, since in case of boron, 2p1 electron has to be removed, to get B+ while in case of Be (2s2), s-electron has to be, removed to get Be+ (2s1). p-electron can be removed, , www.mediit.in

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Classification of Elements and Periodicity in Properties, , 21, , more easily than s-electron so the energy required to, remove electron will be less in case of boron. The order, will be, Li < B < Be < C., 31. (a) : The energy required to remove the most, loosely bound electron from an isolated gaseous atom is, called the ionisation energy., The ionisation potential increases as the size of the atom, decreases. Atoms with fully or partly filled orbitals have, high ionisation potential., 32. (c) : Among the halogens the electron affinity value, of ‘F’ should be maximum. But due to small size there is, inter-electronic repulsion thus, there is difficulty in entry, of new electrons. Thus, the E.A. value is slightly lower, than chlorine and the order is I < Br < F < Cl., 33. (d) : 4Be  1s2 2s2, 5B  1s2 2s2 2p1, Due to stable fully-filled ‘s’-orbital arrangement of, electrons in ‘Be’ atom, more energy is required to, remove an electron from the valence shell than ‘B’-atom., Therefore ‘Be’ has higher ionisation potential than ‘B’., 34. (b) : Positive ion is always smaller and negative ion, is always larger than the parent atom., 35. (d) : Since all of these ions contain 18 electrons, each, so these are isoelectronic. For isoelectronic ions,, the anion having large negative charge is the largest in, size i.e., S2–., 36. (a) : These are isoelectronic ions (ions with same, number of electrons) and for isoelectronic ions, greater, the positive charge, greater is the force of attraction on, the electrons by the nucleus and the smaller is the size of, the ion. Thus, Al3+ has the smallest size., , 37. (b) : Al2O3 and ZnO are amphoteric. N2O5 is, strongly acidic. MgO is the most basic., 38. (c) : The cations are always smaller than the neutral, atom and anions are always larger in size, Na– > Na > Na+., 39. (c) : In isoelectronic ions, the size of the cation, decreases as the magnitude of the positive charge, increases., 40. (d) : Within a period from left to right, atomic, volume first decreases and then increases., 41. (d) : Abnormally high difference between 2nd and, 3rd ionisation energy means that the element has two, valence electrons, which is a case in configuration (d)., 42. (d), 43. (c) : The order of ionic size for given ions will be, K+ > Ca2+ > Mg2+ and that of Cl– > F–. Therefore, MgCl2, has minimum value of cation/anion (Mg2+/Cl–) ratio., 44. (c) : N, O and F are highly electronegative nonmetals and will have the strongest tendency to form, anions by gaining electrons from metal atoms., 45. (c) : As halogens have seven electrons (ns2np5) in, the valence shell, they have a strong tendency to acquire, the nearest inert gas configuration by gaining an electron, from the metallic atom and form halide ions easily., 46. (a) : Metallic character decreases in a period and, increases in a group., 47. (c) : The atomic size decreases within a period from, left to right, therefore Li > Be and Na > Mg. The size, increases in a group from top to bottom. Hence, the size, of Na is greater than Li. Overall order Na > Mg > Li > Be., Thus, Be has smallest size., , , , www.neetujee.com, , www.mediit.in

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, , , , , , , CHAPTER, , Chemical Bonding and, Molecular Structure, , 4, , 4.1, 1., , 2., , 3., , (c) Nitrogen trifluoride, beryllium difluoride, water,, 1, 3-dichlorobenzene, (d) Boron trifluoride, beryllium difluoride, carbon, dioxide, 1, 4-dichlorobenzene, (NEET 2020), , Kössel-Lewis Approach to Chemical, Bonding, , In PO 43– ion, the formal charge on each oxygen atom, and P—O bond order respectively are, (a) –0.75, 1.25, (b) –0.75, 1.0, (c) –0.75, 0.6, (d) –3, 1.25, (1998), Among LiCl, BeCl2, BCl3 and CCl4, the covalent, bond character follows the order, (a) BeCl2 > BCl3 > CCl4 < LiCl, (b) BeCl2 < BCl3 < CCl4 < LiCl, (c) LiCl < BeCl2 < BCl3 < CCl4, (d) LiCl > BeCl2 > BCl3 > CCl4, (1990), Which one of the following formulae does not, correctly represent the bonding capacities of the two, atoms involved?, , Which of the following is the correct order of dipole, moment ?, (a) NH3 < BF3 < NF3 < H2O, (b) BF3 < NF3 < NH3 < H2O, (c) BF3 < NH3 < NF3 < H2O, (d) H2O < NF3 < NH3 < BF3 (Odisha NEET 2019), , 7., , The species, having bond angles of 120° is, (a) ClF3, (b) NCl3, (c) BCl3, (d) PH3, (NEET 2017), , 8., , Consider the molecules CH4, NH3 and H2O. Which, of the given statements is false?, (a) The H — O — H bond angle in H2O is smaller, than the H — N — H bond angle in NH3., (b) The H — C — H bond angle in CH4 is larger, than the H — N — H bond angle in NH3., (c) The H — C — H bond angle in CH4, the, H — N — H bond angle in NH3, and the, H — O — H bond angle in H2O are all greater, than 90°., (d) The H — O — H bond angle in H2O is larger, than the H — C — H bond angle in CH4., (NEET-I 2016), Which of the following molecules has the maximum, dipole moment?, (a) CO2, (b) CH4, (c) NH3, (d) NF3, (2014), , +, , H, , (a) H P, , 6., , H, , (b), , F, , F, O, , H, O, , (c), , (d), , H C C, , OH, , (1990), , 4.2, 4., , Among the following, which compound will show, the highest lattice energy?, (a) KF, (b) NaF, (c) CsF, (d) RbF, (1993), , 4.3, 5., , Ionic or Electrovalent Bond, , Bond Parameters, , Which of the following, set of molecules will have, zero dipole moment?, (a) Ammonia, beryllium difluoride, water,, 1, 4-dichlorobenzene, (b) Boron trifluoride, hydrogen fluoride, carbon, dioxide, 1, 3-dichlorobenzene, , www.neetujee.com, , 9., , 10. The correct order of increasing bond length of, C – H, C – O, C – C and C C is, (a) C – H < C C < C – O < C – C, (b) C – C < C C < C – O < C – H, (c) C – O < C – H < C – C < C C, (d) C – H < C – O < C – C < C C, (2011), 11. Which of the following structures is the most, preferred and hence of lowest energy for SO3?, , www.mediit.in

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Chemical Bonding and Molecular Structure, , 23, , :O:, , 18. H2O is dipolar, whereas BeF2 is not. It is because, (a) the electronegativity of F is greater than that, of O, (b) H2O involves hydrogen bonding whereas BeF2, is a discrete molecule, (c) H2O is linear and BeF2 is angular, (d) H2O is angular and BeF2 is linear., (2004), , :, , S, , (a), , (b), , S, :O:, , :, , :O:, , S, O: :O, O, , S, , (d), , O, , O, , :, , (c), , O, , O, , O, , O, , (Mains 2011), , 12. The correct order of increasing bond angles in the, following triatomic species is, (a) NO + < NO < NO – (b) NO + < NO – < NO, 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , (c) NO – < NO + < NO (d) NO – < NO < NO +, , (2008), , 13. The correct order of C – O bond length among CO,, CO 32–, CO2 is, 2–, , 19. Which of the following molecules does not possess, a permanent dipole moment?, (a) CS2, (b) SO3, (c) H2S, (d) SO2, (1994), 20. The table shown below gives the bond dissociation, energies (Ediss) for single covalent bonds of carbon, (C) atoms with element A, B, C and D. Which, element has the smallest atoms?, , 2–, , Bond, , E diss(kJ mol–1), , C-A, , 240, , (2007), , C-B, , 328, , 14. The electronegativity difference between N and F is, greater than that between N and H yet the dipole, moment of NH3 (1.5 D) is larger than that of NF 3, (0.2 D). This is because, (a) in NH3 the atomic dipole and bond dipole are in, the opposite directions whereas in NF3 these are, in the same direction, (b) in NH3 as well as in NF3 the atomic dipole and, bond dipole are in the same direction, (c) in NH3 the atomic dipole and bond dipole are in, the same direction whereas in NF3 these are in, opposite directions, (d) in NH3 as well as in NF3 the atomic dipole and, bond dipole are in opposite directions. (2006), , C-C, , 276, , C-D, , 485, , (a) CO < CO3 < CO2 (b) CO3 < CO2 < CO, (c) CO < CO 2 < CO 2– (d) CO < CO 2– < CO, 3, , 2, , 3, , 15. The correct order in which the O – O bond length, increases in the following is, (a) O 2 < H2O2 < O3, (b) O 3 < H 2O2 < O2, (c) H 2O2 < O2 < O3, (d) O 2 < O 3 < H2O2, (2005, 1995), 16. The correct sequence of increasing covalent, character is represented by, (a) LiCl < NaCl < BeCl2, (b) BeCl2 < LiCl < NaCl, (c) NaCl < LiCl < BeCl2, (d) BeCl2 < NaCl < LiCl, (2005), 17. Which of the following would have a permanent, dipole moment?, (a) SiF4, (b) SF4, (c) XeF4, (d) BF3, (2005), , www.neetujee.com, , (a) C, (c) A, , (b) D, (d) B, , (1994), , 21. Strongest bond is in between, (a) CsF, (b) NaCl, (c) both (a) and (b), (d) none of the above., (1993), 22. Which of the following bonds will be most polar?, (a) N – Cl, (b) O – F, (c) N – F, (d) N – N, (1992), , 4.4, , The Valence Shell Electron Pair, Repulsion (VSEPR) Theory, , 23. In the structure of ClF3, the number of lone pairs of, electrons on central atom ‘Cl’ is, (a) one, (b) two, (c) four, (d) three, (NEET 2018), 24. Predict the correct order among the following :, (a) bond pair - bond pair > lone pair - bond pair, > lone pair - lone pair, (b) lone pair - bond pair > bond pair - bond pair, > lone pair - lone pair, (c) lone pair - lone pair > lone pair - bond pair, > bond pair - bond pair, (d) lone pair - lone pair > bond pair - bond pair, > lone pair - bond pair, (NEET-I 2016), , www.mediit.in

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24, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 25. Which of the following species contains three bond, pairs and one lone pair around the central atom?, (a) H2O, (b) BF 3, (2012), (c) NH –, (d) PCl, 2, , 3, , 26. Which of the following is not a correct statement?, (a) Multiple bonds are always shorter than, corresponding single bonds., (b) The electron-deficient molecules can act as, Lewis acids., (c) The canonical structures have no real existence., (d) Every AB5 molecule does in fact have square, pyramid structure., (2006), 27. Which of the following is not isostructural with, SiCl4? +, (a) NH, (b) SCl, 4, , (c) SO 2–, 4, , 4, , (d) PO 3–, 4, , (2006), , 28. In which of the following molecules all the bonds, are not equal?, (a) NF3, (b) ClF3, (c) BF3, (d) AlF3, (2006), 29. Which of the following molecules has trigonal, planar geometry?, (a) BF3, (b) NH3, (c) PCl3, (d) IF3, (2005), 30. In a regular octahedral molecule, MX6 the number, of X – M – X bonds at 180° is, (a) three, (b) two, (c) six, (d) four., (2004), 31. In BrF3 molecule, the lone pairs occupy equatorial, positions to minimize, (a) lone pair - bond pair repulsion only, (b) bond pair - bond pair repulsion only, (c) lone pair - lone pair repulsion and lone pair bond pair repulsion, (d) lone pair - lone pair repulsion only., (2004), –, 32. In NO 3 ion, number of bond pair and lone pair of, electrons on nitrogen atom are, (a) 2, 2, (b) 3, 1, (c) 1, 3, (d) 4, 0., (2002), 33. In which of the following bond angle, is maximum?, (a) NH 3, (b) NH+, 4, , (c) PCl3, , (d) SCl2, , (2001), , 34. The BCl3 is a planar molecule whereas NCl3 is, pyramidal because, (a) nitrogen atom is smaller than boron atom, (b) BCl3 has no lone pair but NCl3 has a lone pair of, electrons, (c) B—Cl bond is more polar than N—Cl bond, (d) N—Cl bond is more covalent than B—Cl bond., (1995), , www.neetujee.com, , 35. In compound X, all the bond angles are exactly, 109°28, X is, (a) chloromethane, (b) carbon tetrachloride, (c) iodoform, (d) chloroform., (1991), , 4.5, , Valence Bond Theory, , 36. Which of the following species contains equal, number of  and -bonds?, (a), (b) XeO, (CH),2(CN) 2, (c) (CN), HCO–2, (d), 3, , 4, , (2015, Cancelled), 37. Which one of the following molecules contains no , bond?, (a) SO2, (b) NO2, (c) CO2, (d) H2O, (NEET 2013), 38. Which one of the following statements is not correct, for sigma- and pi- bonds formed between two, carbon atoms?, (a) Sigma-bond is stronger than a pi-bond., (b) Bond energies of sigma- and pi-bonds are of the, order of 264 kJ/mol and 347 kJ/mol, respectively., (c) Free rotation of atoms about a sigma-bond is, allowed but not in case of a pi-bond., (d) Sigma-bond determines the direction between, carbon atoms but a pi-bond has no primary, effect in this regard., (2003), 39. Main axis of a diatomic molecule is z, molecular, orbital px and py overlap to form which of the, following orbitals?, (a)  molecular orbital, (b)  molecular orbital, (c)  molecular orbital, (d) No bond will form., (2001), 40. Which statement is not correct?, (a) A sigma bond is weaker than a pi bond., (b) A sigma bond is stronger than a pi bond., (c) A double bond is stronger than a single bond., (d) A double bond is shorter than a single bond., (1990), 41. Linear combination of two hybridized orbitals, belonging to two atoms and each having one, electron leads to the formation of, (a) sigma bond, (b) double bond, (c) co-ordinate covalent bond, (d) pi bond., , (1990), , 42. Which of the following does not apply to metallic, bond?, (a) Overlapping valence orbitals, (b) Mobile valence electrons, , www.mediit.in

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Chemical Bonding and Molecular Structure, , 25, , (c) Delocalized electrons, (d) Highly directed bonds, , (1989), , (b) NO 3–, (d) CO, , (a) N3 –, (c) NO, 2, , 43. The angle between the overlapping of one s-orbital, and one p-orbital is, (a) 180°, (b) 120°, (c) 109°28, (d) 120°, 60°, (1988), , 4.6, , Hybridisation, , 44. Which of the following pairs of compounds is, isoelectronic and isostructural? –, (a) TeI2, XeF2, (b) IBr , XeF2, 2, (c) IF , XeF, (d) BeCl2, XeF2, 3, 2, (NEET 2017), 45. The hybridizations of atomic orbitals of nitrogen in, NO+2, NO–3 and NH+4 respectively are, (a) sp, sp3 and sp2, (b) sp2, sp3 and sp, 2, 3, (c) sp, sp and sp, (d) sp2, sp and sp3, (NEET-II 2016), 46. Which of the following pairs of ions is isoelectronic, and isostructural?, (a) CO 2–, NO –, (b) ClO – , CO 2–, 3, , 3, , (c) SO2– , NO –, 3, , 3, , 3, , 3, , 3, , (NEET-II 2016), 47. The correct geometry and hybridization for XeF 4 are, (a) octahedral, sp3d2, (b) trigonal bipyramidal, sp d, (c) planar triangle, sp3d3, (d) square planar, sp3d2., , (NEET 2013), , 53. Which of the following is a polar molecule?, (a) SiF4, (b) XeF4, (c) BF3, (d) SF4, (NEET 2013), 54. The outer orbitals of C in ethene molecule can be, considered to be hybridized to give three equivalent, sp2 orbitals. The total number of sigma () and pi, () bonds in ethene molecule is, (a) 3 sigma () and 2 pi () bonds, (b) 4 sigma () and 1 pi () bonds, (c) 5 sigma () and 1 pi () bonds, (d) 1 sigma () and 2 pi () bonds., (Karnataka NEET 2013), 55. In which of the following pairs both the species have, sp3 hybridization?, (a) SiF4, BeH2, (b) NF3, H2O, (c) NF3, BF3, (d) H2S, BF3, (Karnataka NEET 2013), 56. Which one of the following pairs is isostructural, (i.e., having the same shape and hybridization)?, (a) [BCl3 and BrCl3], (b) [NH3 and NO–], 3, , –, , (c) [NF3 and BF3], , 3, , (d) [BF4 and NH +4], (2012), , (NEET-II 2016), , 48. Among the following, which one is a wrong, statement?, (a) PH5 and BiCl5 do not exist., (b) p-d bonds are present in SO2., (c) SeF4 and CH4 have same shape., (d) I + has bent geometry., (NEET-II 2016), 3, , 49. In which of the following pairs, both the species are, not isostructural?, (a) Diamond, Silicon carbide, (b) NH 3, PH3, (c) XeF4, XeO4, (d) SiCl4, PCl4+, (2015), 50. Maximum bond angle at nitrogen is present in, which of the following?, (a) NO2+, (b) NO3–, (c) NO 2, (d) NO 2–, (2015, Cancelled), 51. Which one of the following species has planar, triangular shape?, , www.neetujee.com, , 52. XeF2 is isostructural with, (a) SbCl3, (b) BaCl2, (c) TeF2, (d) ICl2–, , (d) ClO –, SO2–, , 3, , (2014), , 2, , 57. Which of the two ions from the list given below that, have the geometry that is explained, by the same, hybridization of orbitals, NO –, NO –, NH –, NH +,, 2, 3, 2, 4, SCN– ?, –, –, +, –, (a) NO and NO, (b) NH and NO, 2, , 3, , (c) SCN– and NH 2–, , 4, , 3, , 2, , 2, , (d) NO – and NH – (2011), , 58. In which of the following pairs of molecules/ions,, the central, atoms have sp2 hybridisation? –, (a) NO2– and NH3, (b) BF3and NO 2, (c) NH – and H O, (d) BF and NH – (2010), 2, , 2, , 3, , 2, , 59. In which one of the following species the central, atom has the type of hybridization which is not the, same as that present in the other, three?, (a) SF4, (b) I –, 3, (2010), (d) PCl5, 60. In which of the following molecules the central, atom does not have sp3 hybridization?, (a) CH–4, (b) SF4 +, (c) BF, (d) NH, (Mains 2010), , (c) SbCl52–, , 4, , 4, , www.mediit.in

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26, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , NO3–, , 61. Some of the properties of the two species,, and, H3O+ are described below. Which one of them is, correct?, (a) Dissimilar in hybridization for the central atom, with different structures., (b) Isostructural with same hybridization for the, central atom., (c) Isostructural with different hybridization for, the central atom., (d) Similar in hybridization for the central atom, with different structures., (Mains 2010), 62. In which of the following molecules/ions BF3, NO2–,, NH 2– and H 2O, the central atom is sp2 hybridised?, –, –, (a) NH 2 and H2O, (b) NO 2 and H2O, (c) BF 3 and NO2–, , (d) NO –2 and NH 2– (2009), , 63. In which of the following pairs, the two species are, isostructural?, (a) SO 2– and NO –, (b) BF and NF, 3, , (c), , –, BrO3, , 3, , and XeO3, , 3, , 3, , (d) SF4 and XeF4, , (2007), , 64. Which of the following species has a linear shape?, (a) O3, (b) NO –2, (c) SO2, (d) NO2+, (2006), 65. The correct order regarding the electronegativity of, hybrid orbitals of carbon is, (a) sp < sp2 < sp3, (b) sp > sp2 < sp3, (c) sp > sp2 > sp3, , (d) sp < sp2 > sp3, , (2006), , 66. Among the following, the pair in which the two, species are not isostructural is, (a) SiF4 and SF4, (b) IO 3– and XeO3, –, +, (c) BH and NH, (d) PF – and SF . (2004), 4, , 4, , 6, , (d) CO 2–, , 3, , (2002), , 3, , 2, , (c) CO3 , SO3, , 2–, , 3, , 3, , (1992), , 4, , 74. A sp3 hybrid orbital contains, (a) 1/4 s-character, (b) 1/2 s-character, (c) 1/3 s-character, , (d) 2/3 s-character., , 75. The complex ion [Co(NH3)6], , 3+, , (1991), 32, , is formed by sp d, , hybridisation. Hence the ion should possess, (a) octahedral geometry, (b) tetrahedral geometry, (c) square planar geometry, (d) tetragonal geometry., (1990), 76. Which of the following molecules does not have a, linear arrangement of atoms?, (a) H2S, (b) C2H2, (c) BeH2, (d) CO2, (1989), 77. In which one of the following molecules the central, atom can be said to adopt sp2 hybridization?, (a) BeF2, (b) BF3, (1989), (c) C H, (d) NH, 2, , 78. Equilateral shape has, (a) sp3hybridisation, (c) sp hybridisation, , 3, , (b) sp2 hybridisation, (d) dsp3 hybridisation., (1988), , 3, , (d) PCl5, ICl5, , (2001), , 69. The bond length between hybridised carbon atom, and other carbon atom is minimum in, (a) propene, (b) propyne, (c) propane, (d) butane., (1996), 70. Which of the following has sp2-hybridisation?, (a) BeCl2, (b) C2H2, (c) C2H6, (d) C2H4, (1996), , www.neetujee.com, , 73. Which structure is linear?, (a) SO2, (b) CO2, (c) CO 2–, (d) SO 2–, , 2, , 68. Which of the following two are isostructural?, (a) XeF2, IF –, (b) NH , BF, 2–, , 72. Which one of the following has the shortest carbon, carbon bond length?, (a) Benzene, (b) Ethene, (c) Ethyne, (d) Ethane, (1992), , 6, , 67. Which of the following has p – d bonding?, (a) NO3–, (b) SO 32–, (c) BO 3–, , 71. When the hybridization state of carbon atom, changes from sp3 to sp2 and finally to sp, the angle, between the hybridized orbitals, (a) decreases gradually, (b) decreases considerably, (c) is not affected, (d) increases progressively., (1993), , 4.7, , Molecular Orbital Theory, , 79. Consider the following species : CN+, CN –, NO and, CN. Which one of these will have the highest bond, order?, (a) NO, (c) CN+, , (b) CN–, (d) CN, , (NEET 2018), , 80. Which one of the following pairs of species have the, same bond order?, (a) O2, NO+, (b) CN–, CO, –, (c) N2, O2, (d) CO, NO (NEET 2017), , www.mediit.in

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Chemical Bonding and Molecular Structure, , 27, , 81. Which of the following is paramagnetic?, (a) CN –, (b) NO+, (NEET 2013), (c) CO, (d) O 2–, 82. The pair of species that has the same bond order in, the following is, (a) CO, NO+, (b) NO –, CN –, (c) O2, N2, (d) O2, B2, (Karnataka NEET 2013), 83. In which of the following ionization processes the, bond energy increases and the magnetic behaviour, changes from paramagnetic to diamagnetic?, (a) O  O +, (b) C  C +, 2, , 2, , (c) NO  NO, , +, , 2, , 2, N2  N2+, , (d), (Karnataka NEET 2013), , 84. The pair of species with the same bond order is, (a) O 2–, B, (b) O +, NO +, 2, , 2, , 2, , which one of the following orbitals?, (a) * orbital, (b)  orbital, (c) * orbital, , (d)  orbital (Mains 2012), , (c) C 2– < He + < O – < NO, 2, 2, 2, (d) He2+ < O – < NO < C2–, 2, , (Mains 2012, 2008), , 2, , 87. Which of the following is isoelectronic?, (a) CO 2, NO2, (b) NO –2, CO2, (c) CN–, CO, (d) SO2, CO2, (2002), 88. Which species does not exhibit paramagnetism?, –, , +, , (b) O2, (d) NO, , (2000), , (1998), , (d) 4, , 90. Which of the following species is paramagnetic?, (a) CO, (b) CN–, (c) O 2–, (d) NO, (1995), 2, , Bonding in Some Homonuclear, Diatomic Molecules, , 91. Identify a molecule which does not exist., (a) He, (b) Li, (NEET 2020), (c) C 2, (d) O 2, 2, , www.neetujee.com, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , (c) O – > O2– > O + > O, 2, , 2, , (d) O + > O > O – > O2–, 2, , (2015), , 2, , 95. The correct bond order in the following species is, (a) O + < O – < O 2+, (b) O –2 < O2+ < O 22+, 22+, 2+, 2–, (c) O < O < O, (d) O 2+ < O – < O +, 2, , 2, , 2, , 2, , 2, , (2015, Cancelled), 96. Bond order of 1.5 is shown by, (a) O2+, (b) O–2, (c) O22–, (d) O2, (2012), 97. Which of the following has the minimum bond, length?, +, –, (a) O2, (b) O2, (c) O22–, (d) O2, (2011), 98. The pairs of species of oxygen and their magnetic, behaviour are noted below. Which of the following, presents the correct description?, (a) O2– , O22– - Both diamagnetic, (b) O +, O 2– - Both paramagnetic, 2, , 89. The number of anti-bonding electron pairs in O22–, molecular ion on the basis of molecular orbital, theory is (Atomic number of O is 8.), (a) 3, (b) 2, , 4.8, , 2, , (a) O22–> O2 >– O2 > O2 +, (b) O > O+ > O2– > O –, , 2, , 86. Four diatomic species are listed below. Identify the, correct order in which the bond order is increasing, in them., (a) NO < O – < C 2– < He +, 2, 2, 2, (b) O2– < NO < C22– < He2+, , (c) 5, , 93. Which of the following is paramagnetic?, (a) N2, (b) H2, (c) Li2, (d) O2, (Odisha NEET 2019), 94. Decreasing, order, of, stability, of, O2, O – , O2 + and, 2, 2–, O is, , 2, , (c) NO, CO, (d) N2, O2, (2012), 85. During change of O2 to O – ion, the electron adds on, , (a) N2, (c) CO, , 92. Which of the following diatomic molecular species, has only  bonds according to Molecular Orbital, Theory?, (a) Be2, (b) O2, (c) N2, (d) C2, (NEET 2019), , (c), , +, O2 ,, , O2 - Both paramagnetic, , (d) O, O22– - Both paramagnetic, , (2011), , 99. Which one of the following species does not exist, under normal, conditions?, (a) Be+, (b) Be, 2, , (c) B2, , 2, , (d) Li2, , (2010), , 100. According to MO theory which of the lists ranks the, nitrogen, species, in terms of increasing, bond, order?, 2–, –, 2–, –, (a) N2 < N2 < N2, (b) N2 < N2 < N2, (c) N – < N 2– < N, (d) N – < N < N 2–, 2, 2, 2, 2, 2, 2, (2009), +, 101. Right order of dissociation energy N2 and N2 is, (a) N > N +, (b) N = N+, 2, 2, 2, 2, (2000), +, (c) N 2 > N2, (d) none., , www.mediit.in

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28, 102. N2 and O2 are converted into monocations, N + and, 2, O+2 respectively. Which is wrong?, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, (a) H2O2, (b) HCN, (c) Cellulose, , +, , (a) In O 2 paramagnetism decreases., (b) N + becomes diamagnetic., , (NEET-II 2016), , (d) Concentrated acetic acid, , 2, , (c) In N+ 2 , the N–N bond weakens., (d) In O+2, the O–O bond order increases. (1997), 103. N2– and O2 are converted into monoanions N2– and, O2 respectively, which of the following statements is, wrong?, (a) In–O–2, bond length increases., (b) N2 becomes diamagnetic., (c) In N –2 , N–N bond weakens., (d) In O–2, the O–O bond order decreases. (1997), 104. The ground state electronic configuration of valence, shell electrons, in nitrogen molecule (N2) is written, as KK, 2s2, *2s2, 2p 2 = 2p 22p 2. Hence the, x, , y, , z, , bond order in nitrogen molecule is, (a) 2, (b) 3, (c) 0, (d) 1, , (c), , 4.9, , 2, , O2+, , (d) O22–, , (b) Covalent bonds, (c) Hydrogen, London dispersion, (d), bondingforce, , (c) in both electron density will decrease, (d) on X electron density will decrease and on H, increases., (2001), 109. Strongest hydrogen bond is shown by, (a) water, (b) ammonia, (c) hydrogen fluoride, (d) hydrogen sulphide., , (1994), , Hydrogen Bonding, , 106. Which one of the following compounds shows the, presence of intramolecular hydrogen bond?, , (2009), , 108. In X – H, Y, X and Y both are electronegative, elements. Then, (a) electron density on X will increase and on H, will decrease, (b) in both electron density will increase, , (1995), , 105. Which of the following molecules has the highest, bond order?, (a) O –, (b) O, 2, , 107. What is the dominant intermolecular force or bond, that must be overcome in converting liquid CH3OH, to, gas?, (a)aDipole-dipole, interaction, , (1992), , 110. Which one shows maximum hydrogen bonding?, (a) H2O, (b) H2Se, (c) H2S, (d) HF, (1990), , ANSWER KEY, , 1., 11., 21., 31., 41., 50., 60., 70., 80., 90., 100., 110., , (a), (d), (a), (d), (a), (a), (b), (d), (b), (d), (a), (a), , www.neetujee.com, , 2., 12., 22., 32., 42., 51., 61., 71., 81., 91., 101., , (c), (d), (c), (d), (d), (b), (a), (d), (d), (a), (a), , 3., 13., 23., 33., 43., 52., 62., 72., 82., 92., 102., , (d), (c), (b), (b), (a), (d), (c), (c), (a), (d), (b), , 4., 14., 24., 34., 44., 53., 63., 73., 83., 93., 103., , (b) 5., (c) 15., (c) 25., (b) 35., (None), (d) 54., (c) 64., (b) 74., (c) 84., (d) 94., (b) 104., , (d), (d), (d), (b), (c), (d), (a), (a), (d), (b), , 6., 16., 26., 36., 45., 55., 65., 75., 85., 95., 105., , (b), (c), (d), (d), (c), (b), (c), (a), (a), (b), (c), , 7., 17., 27., 37., 46., 56., 66., 76., 86., 96., 106., , (c), (b), (b), (d), (a,d), (d), (a), (a), (d), (b), (c), , 8., 18., 28., 38., 47., 57., 67., 77., 87., 97., 107., , (d), (d), (b), (b), (a), (a), (b), (b), (c), (a), (d), , 9., 19., 29., 39., 48., 58., 68., 78., 88., 98., 108., , (c), (a), (a), (a), (c), (b), (a), (b), (c), (c), (a), , 10., 20., 30., 40., 49., 59., 69., 79., 89., 99., 109., , (a), (b), (a), (a), (c), (c), (b), (b), (d), (b), (c), , www.mediit.in

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Chemical Bonding and Molecular Structure, , 29, , Hints & Explanations, 1. (a) : The total charge = –3, So, the average formal charge on each ‘O’ atom is –3/4, = –0.75, O–, O, O–, O–, –, , P, , P, , –, , O O– O, , O, , –, , P, , –, , –O, , O, , –, , O, , –, , 7. (c) : BCl3-Trigonal planar, sp2-hybridised, 120°, angle., H, 8., , (d) :, , P, , O O–, , O, O,  Average P—O bond order, Total no. of bonds, 5, ,   1.25, Total no. of resonating structures 4, , H, , O, , 109°28, , C, , N, H, , HH, , O, , 107°, , H, , H, , H, , 104.5°, , H, , H, , 9., , (c) : O, , C, , C, , O ;, , ( = 0, symmetrical), , 2. (c) : Along the period, as we move from, Li  Be  B  C, the electronegativity increases, and hence the EN difference between the element and, Cl decreases and accordingly, the covalent character, increases. Thus LiCl < BeCl2 < BCl3 < CCl4 is the correct, order of covalent bond character., , H, , H, , H, , ( = 0, symmetrical), , N, , N, , H, , H, , F, , H, , F, , F, , (  0, pyramidal), , (  0, pyramidal), , The asterisk (*) marked carbon has a valency of 5 and, hence, this formula is not correct because carbon has a, maximum valency of 4., 4. (b) : For compounds containing ions of same charge,, lattice energy increases as the size of ions decreases. Thus,, NaF has highest lattice energy., F, , In NH3, H is less electronegative than N and hence, dipole moment of each N—H bond is towards N and, create high net dipole moment whereas in NF3, F is more, electronegative than N, the dipole moment of each N—F, bond is opposite to that of lone pair, hence reducing the, net dipole moment., 10. (a) : Increasing order of bond length is, C— H < C C < C—O < C— C, O, , 5., , 11. (d) :, , 3., , O, C* O, , (d) : H C, , (d) : F, , B, , H, , F, F, , Boron trifluoride, ( = 0), , Be, , F, , Beryllium difluoride, ( = 0), , Cl, O, , C, ,  , ,  S  has maximum number of covalent, OO, bonds involving p – d bonding also., , 12. (d) : Structures of NO–2, NO2 and NO+2 is given as, , O, , Carbon dioxide, ( = 0), , –, , 13. (c) : CO  . C, , (b) :, , CO 32– , O–, , .C, , O, , O, , O, , C, , C, , C, , O– O–, C, , O., , O, , O, +, , .O, , www.neetujee.com, , O., , –, O, –, O.., , C, , .O, , C, , –., , .., , 1.85 D, , ., , CO2  . O, , +, , O., , ., , 6., , The correct order of increasing bond angles in the, following triatomic species is, NO2– < NO 2 < NO2 +, , .., , Cl, 1, 4-Dichlorobenzene, ( = 0), , +, , ., , O, , www.mediit.in

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30, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , More single bond character in resonance hybrid, more, is the bond length. Hence, the increasing bond length is, CO < CO2 < CO32–, 14. (c) : The dipole moment of NF3 is 0.24 D and, of NH3 is 1.48 D. The difference is due to fact that the, dipole moment due to N – F bonds in NF3 are in opposite, directions to the direction of the dipole moment of the, lone pair on N atom which partly cancel out. The dipole, moment of N – H bonds in NH3 are in the same direction, of the dipole moment of the lone pair on N atom which, adds up as shown :, .., , .., , N, , N, F, , F, , F, NF3, , H, , H, , H, NH3, , 15. (d) : Bond lengths of O – O in O2 is 1.21 Å, in H2O2, is 1.48 Å and in O3 is 1.28 Å. Therefore, correct order of, the O – O bond length is H2O2 > O3 > O2., 16. (c) : Covalent character in a compound is found by, Fajan’s rule., Fajan’s rule : The smaller the size of the cation and the, larger the size of the anion, the greater is the covalent, character of an ionic bond. The greater the charge on the, cation, the greater is the covalent character of the ionic, bond., 17. (b) : For dipole moment, we have to know the, hybridisation and shape., , O, Be, , F, H, ,  ≠0, , H, , F, , =0, , 19. (a) : The structure of CS2 is linear and therefore, it does not have permanent dipole moment. It is, represented as S C S., 20. (b) : Smaller the atom, stronger is the bond and, greater is the bond dissociation energy. Therefore, the, bond C-D has the greatest energy or D has smallest, atoms., 21. (a) : According to Fajans rule, ionic character, increases with increase in size of the cation,, (Cs > Rb > K > Na) and with decrease in size of the anion, (F > Cl > Br > I). Thus, CsF has higher ionic character, than NaCl and hence, bond in CsF is stronger than in, NaCl., 22. (c) : Polarity of the bond depends upon the, electronegativity difference of the two atoms forming the, bond. Greater the electronegativity difference, more is, the polarity of the bond., N – Cl, O–F N–F, N–N, 3.04–3.16 3.5–4.0 3.04–4.0 3.04–3.04, 23. (b) : The structure of ClF3 is, F, Cl, , F, , F, Hence, Cl has 2 lone pairs of electrons., 24. (c) : According to VSEPR theory, the repulsive, forces between lone pair and lone pair are greater than, between lone pair and bond pair which are further, greater than bond pair and bond pair., 25. (d) :, , F, , 2 bond pairs, 2 lone pairs, , B, , F, , 3 bond pairs, 0 lone pair, , F, N, 18. (d) : The overall value of the dipole moment of a, polar molecule depends on its geometry and shape, i.e., vectorial addition of dipole moment of the constituent, bonds. Water has angular structure with bond angle 105°,, it has dipole moment. However BeF2 is a linear molecule, thus, dipole moment summation of all the bonds present, in the molecule cancel each other., www.neetujee.com, , H, , H, P, , Cl, , Cl, , 2 bond pairs, 2 lone pairs, , Cl 3 bond pairs, 1 lone pair, , 26. (d) : For AB5 molecules, there are three possible, geometries i.e. planar pentagonal, square pyramidal and, trigonal bipyramidal., www.mediit.in

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Chemical Bonding and Molecular Structure, B, , B, , 72°, , B, , B, , A, B, , B, , B, , B, , B, , B B, , A, , 31, , 90°, , 90°, , A, , B, , B, , B, , planar pentagonal square pyramidal, , B, 120°, , –, , 32. (d) : N O, , 4, , 4, , O, , O, , trigonal bipyramidal, , Out of these three geometries, it is only trigonal, bipyramidal shape in which bond pair-bond pair, repulsions are minimum and hence, this geometry is the, most probable geometry of AB5 molecule., 27. (b) : SiCl4, NH+, SO2– and PO 3– ions are the, 4, , Bent T-shaped geometry in which both lone pairs occupy, the equatorial positions of the trigonal bipyramid. Here, (lp - lp) repulsion = 0, (lp - bp) repulsion = 4 and (bp - bp), repulsion = 2., , NO 3–, , In, ion, nitrogen has 4 bond pairs of electrons and, no lone pair of electrons., +, 33. (b) : Bond angle is maximum in NH4, molecule with bond angle 109°., , tetrahedral, , 34. (b) : There is no lone pair on boron in BCl 3 hence,, no repulsion takes place. There is a lone pair on nitrogen, , examples of molecules/ions which are of AB4 type and, have tetrahedral structures. SCl 4 is AB4 (lone pair) type, species. Although the arrangement of five sp3d hybrid, orbitals in space is trigonal bipyramidal, due to the, presence of one lone pair of electrons in the basal hybrid, orbital, the shape of AB4 (lone pair) species gets distorted, and becomes distorted tetrahedral or see-saw., , in NCl3 hence, repulsion takes place. Therefore, BCl3 is, planar molecule but NCl 3 is pyramidal molecule., 35. (b) : As all C – Cl bonds are directed towards the, corner of a regular tetrahedron., 36. (d) : (CN)2, N C — C N (3  + 4 ), , 28. (b) : The Cl – F (Cl – Feq) bond length is equal to, 1.60 Å while each of the two axial Cl – F (Cl – Fa) bond, length is equal to 1.70 Å., , CH2(CN)2,, , (6  + 4 ), , HCO3– ,, , (4  + 1 ), , XeO4,, , (4  + 4 ), , 29. (a) :, F, , F, , ··, , B, , N, H, , H, , F, trigonal planar, , H, pyramidal,  O , 37. (d) : H, H, 38. (b) : Sigma bond dissociation energy = 347 kJ/mol, Pi-bond dissociation energy = 264 kJ/mol, 39. (a) : For  overlap, the lobes of the atomic orbitals, are perpendicular to the line joining the nuclei., , 30. (a) : In octahedral molecule, six hybrid orbitals, directed towards the corners, X, 90°, of a regular octahedron with a, X, bond angle of 90°., X, According to this geometry, 90°, M, the number of X – M – X X, X, bonds at 180° must be three., X, F, 31. (d) :, , Br, , F, , Hence, only sidewise overlapping takes place., 40. (a) : A -bond is stronger than a -bond., 41. (a), 42. (d) : Metallic bonds have electrostatic attractions on, all sides and hence, do not have directional characteristics., 43. (a) : The type of overlapping between s - and, p-orbitals occurs along internuclear axis and hence, the, angle is 180°., , F, , www.neetujee.com, , www.mediit.in

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Chemical Bonding and Molecular Structure, 58. (b) : The hybridisation of the central atom can be, calculated as, No. of electrons No. of monovalent  2, ,  , in valence shell   atoms around, ,  , , 1,   central atom, , H    of atom, , 2, Charge on Charge on, , , ,    anion, ,  cation, , , , , , , 1, For BF ,, [(3)  (3)  (0)  (0)], 3H, =2 3  sp2 hybridisation., 1, For NO – , H  [(5)  (0)  (0)  (1)], 2, 2, = 3  sp2 hybridisation., 59. (c) : Hybridisation of the central atom can be, calculated as:, No. of valence  No. of monovalent  , ,  , electrons in the  atoms around, , ,  centralatom, atom, , H  1central,  , , 2, ,  Charge on  Charge on, , ,  cation,   anion, , , , , Applying this formula we find that all the given, species except [SbCl5]2– have central atom with sp3d, (corresponding to H = 5) hybridization. In [SbCl 5]2–, Sb, is sp3d2 hybridized., 60. (b) : For neutral molecules,, No. of electron pairs = No. of atoms bonded to it +, 1/2[Gp. no. of central atom – Valency of central atom], 1,  For CH4, no. of e– pairs  4  [4  4], = 4 (sp3 hybridisation) 2, 1, –, For SF4, no. of e pairs  4  [6  4], 2, = 5 (sp3d hybridisation), For ions,, No. of electron pairs = No. of atoms bonded to it +, 1/2[Gp. no. of central atom – Valency of central atom ±, No. of electrons equals1to the units of charge],  For BF –, no. of e– pairs  4  [3  4 1], 4, 2, = 4 +(sp3 hybridisation), 1, –, For NH , no. of e pairs  4  [5  4 1], 4, , 2, = 4 (sp3 hybridisation), 61. (a) : No. of electron pairs at the central atom, = No. of atoms bonded to it + 1/2[Group number of central, atom – Valency of the central atom ± No. of electrons, equals to the units of charge], –, No. of electron pairs at the central atom in NO3, 1,  3  [5  6 1]  3 (sp2 hybridisation)., 2, , www.neetujee.com, , 33, , No. of electron pairs at the central atom in, 1, H3O+  3  [6  3 1]  4 (sp3 hybridisation)., 62. (c) : BF3  sp2, NO 2–  sp2, NH2 –  sp3, H2O  sp3, 63. (c) : Hybridisation of Br in BrO– :, 3, , H = 1/2(7 + 0 – 0 + 1) = 4 i.e. sp3 hybridisation, Hybridisation of Xe in XeO3 :, 1, H  (8  0  0  0)  4 i.e. sp3 hybridisation, 2, In both BrO3– and XeO3 , the central atom is sp3 hybridised, and contains one lone pair of electrons, hence in both the, cases, the structure is trigonal pyramidal., .., , .., Br, O, , O, , Xe, , O–, , O, , O, , O, , –, , XeO3, , BrO3, , 64. (d) : NO –2 : Due to sp2 hybridisation of, N, N-atom and the presence of one lone pair, O 115° O, –, on it, NO – has angular shape., 2, , O, , O3 : O, , 116.8° O, , V-shaped, , of one lone pair of electrons, SO : Due to the presence, 2, 2, in, one of the three sp -hybrid orbitals, SO2 molecule has, angular (V-shaped) structure., , S, O 119.5° O, , NO + : Due to sp hybridisation of N+, NO+ ion has linear, 2, , 2, , shape., , +, , O, , N, , O, , 65. (c) : Electronegativity of carbon atom is not, fixed. It varies with the state of hybridisation., Electronegativity of carbon increases as the scharacter of the hybrid orbital increases., C (sp) > C (sp2) > C (sp3), 66. (a) : SiF has symmetrical tetrahedral shape which, 4, , is due to, hybridisation of the central silicon atom., SF4 has distorted, tetrahedral or see-saw geometry which, 3, s, parises due to sp3d hybridisation of central sulphur atom, and due to the presence of one lone pair of electrons in, one of the equatorial hybrid orbital., 67. (b) : In sulphite ion, the central atom sulphur is sp3, hybridised., Electronic structure of S atom in excited state, 3p, 3d, 3s, , www.mediit.in

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36, 95. (b) : O – < O +, 2, , B.O. :, , 1.5, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, N 2– (16) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p )2 (2p )2, , < O2+, , 2, , 2, , 2.5, ,  B.O. , , 96. (b) : Configuration of O2 (16) :, 1s2 *1s2 2s2 *2s2 2p2 2p 2 2p 2 *2p 1 *2p1,  z, , No. of e in, Bond order =, , x, , y, , x, , No. of e in, , 10  5, , 2, – 10, , Bond order of O2 , , Bond order of O 2– , ,  2.5, , 2, 7, , 2, 10  8, , 10  6, 2, 2, , 2–, , 101. (a) : N2(14) : (1s)2, (*1s)2, (2s)2, (*2s)2, (2px)2, (2py)2, (2pz)2, N  Na 10  4, In N2, bond order  b, , 3, 2, 2, 94, In N+, bond order , 25, , 2, ,  1.5, , 2, , 2 is more than N + so the, As the bond order in N, 2, 2, dissociation energy of N is higher than N+., ,  1.0, , 2, 10  6, 2, Bond order of O2 , 2, 97. (a) : Electronic configuration, 2, , 2, , 2, , 2, , 2, , 1, , 1, , electron in *2px orbital., 104. (b) : Number of electrons in bonding orbitals, Nb = 10 and number of electrons in antibonding orbitals, Na = 4., Therefore bond order = 1/2(Nb – Na) = 1/2(10 – 4) = 3, 105. (c) : The bond order of O + = 2.5, O 2– = 1,, 2, O2– = 1.5 and that of O 2 = 2. 2, , 2, , 1, , : Bond order = (8  6)  1, 2, As bond order increases, bond length decreases., , O, , 2–, , 2, , 102. (b) : Diamagnetism is caused due to the absence of, unpaired electrons. But in N2 +, there is unpaired electron., So, it is paramagnetic., 103. (b) : N– becomes paramagnetic due to one unpaired, 2, , O2 : KK(2s) (*2s) (2pz) (2px) (2py) (*2px) (*2py), 1, Bond order = (8  4)  2, 2, O1: Bond order, 1+ = (8  3)  2, 2, 2, 2, 1, 1, O– : Bond order = 2 (8  5)  1, 2, , –, , Hence, bond order increases as : N2 < N2 < N2, , 2, , 2, , y, , (2pz)2 (*2px)1 (*2py)1, , y, , , bonding M.O.  antibonding M.O., , Bond order of O + , , x, , 2, , 3.0, , 2, , 98. (c) : O +2 and O2 are paramagnetic in nature as they, contain one and two unpaired electrons respectively., 99. (b) : Be2 does not exist., Be2 has an electronic configuration of :, 1s2 *1s2 2s2 *2s2, 44,  Bond order =, 0, 2, Thus, Be2 does not exist., 100. (a) : According to MOT, the molecular orbital, electronic configuration of, N2 (14) : (1s)2 (*1s)2 (2s)2 (*2s)2(2px)2 (2py)2 (2pz)2, 10  4,  B.O , 3, 2, 2 1, N2– (15) : (1s)2 (*1s)2 (2s)2 (*2s) 2 (2px()2p(2p, )2 (, y )p ), 2 z, *2 x, 10  5,  B.O ,  2.5, 2, , 106. (c) : H2O2, HCN and conc. CH3COOH form, intermolecular hydrogen bonding while cellulose has, intramolecular hydrogen bonding., 107. (d) : Methanol can undergo intermolecular, association through H-bonding as the – OH group in, alcohols is highly polarised., CH3, , CH3, , CH3, , ---O—H---O—H---O—H---, , As a result, in order to convert liquid CH 3OH to gaseous, state, the strong hydrogen bonds must be broken., 108. (a) : –X — H+ ........... Y, the electrons of the covalent, bond are shited towards the more electronegative atom., This partially positively charged H-atom forms hydrogen, bond with the other more electronegative atom., 109. (c) : H – F shows strongest H-bonds because, fluorine, electronegative., 110. (a) :isHmost, O shows, maximum H-bonding because, 2, , each H2O molecule is linked to four H2O molecules, through H-bonds., , , , , , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 5, , 5.1, 1., , 2., , States of Matter, (a) 0°C, (c) absolute zero, , Intermolecular Forces, , Dipole-induced dipole interactions are present in, which of the following pairs?, (a) HCl and He atoms (b) SiF4 and He atoms, (c) H2O and alcohol, (d) Cl2 and CCl4, (NEET 2013), Which one of the following is the correct order of, , 5.6, , 5.4, 3., , Which of the following statements is wrong for, gases?, (a) Confined gas exerts uniform pressure on the, walls of its container in all directions., (b) Volume of the gas is equal to volume of container, confining the gas., (c) Gases do not have a definite shape and volume., (d) Mass of a gas cannot be determined by weighing, a container in which it is enclosed., (1999), , 5.5, 4., , 5., , The Gaseous State, , A mixture of N2 and Ar gases in a cylinder contains, 7 g of N2 and 8 g of Ar. If the total pressure of the, mixture of the gases in the cylinder is 27 bar, the, partial pressure of N2 is, [Use atomic masses (in g mol–1) : N = 14, Ar = 40], (a) 9 bar, (b) 12 bar, (c) 15 bar, (d) 18 bar. (NEET 2020), , 7., , The volume occupied by 1.8 g of water vapour at, 374°C and 1 bar pressure will be, [Use R = 0.083 bar L K–1 mol–1], (a) 96.66 L, (b) 55.87 L, (c) 3.10 L, (d) 5.37 L, (Odisha NEET 2019), , 8., , Equal moles of hydrogen and oxygen gases are, placed in a container with a pin-hole through which, both can escape. What fraction of the oxygen escapes, in the time required for one-half of the hydrogen to, escape?, (a) 3/8, (b) 1/2, (c) 1/8, (d) 1/4, (NEET-I 2016), , 9., , What is the density of N2 gas at 227°C and 5.00 atm, pressure? (R = 0.082 L atmK–1 mol–1), (a) 1.40 g/mL, (b) 2.81 g/mL, (c) 3.41 g/mL, (d) 0.29 g/mL, (Karnataka NEET 2013), , 10., , 50 mL of each gas A and of gas B takes 150 and 200, seconds respectively for effusing through a pin hole, under the similar conditions. If molecular mass of, gas B is 36, the molecular mass of gas A will be, (a) 96, (b) 128, (c) 32, (d) 64, (2012), , 11., , A certain gas takes three times as long to effuse out, as helium. Its molecular mass will be, , The Gas Laws, , At 25°C and 730 mm pressure, 380 mL of dry oxygen, was collected. If the temperature is constant, what, volume will the oxygen occupy at 760 mm pressure?, (a) 569 mL, (b) 365 mL, (c) 265 mL, (d) 621 mL, (1999), Pressure remaining the same, the volume of a given, mass of an ideal gas increases for every degree, centrigrade rise in temperature by definite fraction, of its volume at, , www.neetujee.com, , Ideal Gas Equation, , 6., , interactions?, (a) Covalent < hydrogen bonding < van der Waals’, < dipole-dipole, (b) van der Waals’ < hydrogen bonding < dipoledipole < covalent, (c) van der Waals’ < dipole-dipole < hydrogen, bonding < covalent, (d) Dipole-dipole < van der Waals’ < hydrogen, bonding < covalent., (1993), , (b) its critical temperature, (d) its Boyle temperature., (1989), , www.mediit.in

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38, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (a) 27 u, (c) 64 u, 12., , 13., , (b) 36 u, (d) 9 u, , (Mains 2012), , Two gases A and B having the same volume diffuse, through a porous partition in 20 and 10 seconds, respectively. The molecular mass of A is 49 u., Molecular mass of B will be, (a) 50.00 u, (b) 12.25 u, (c) 6.50 u, (d) 25.00 u, (2011), A gaseous mixture was prepared by taking equal, moles of CO and N2. If the total pressure of the, mixture was found 1 atmosphere, the partial, pressure of the nitrogen (N 2) in the mixture is, (a) 0.5 atm, (b) 0.8 atm, (c) 0.9 atm, (d) 1 atm, (2011), , 14. A bubble of air is underwater at temperature 15°C, and the pressure 1.5 bar. If the bubble rises to the, surface where the temperature is 25°C and the, pressure is 1.0 bar, what will happen to the volume, of the bubble?, (a) Volume will become greater by a factor of 1.6., (b) Volume will become greater by a factor of 1.1., (c) Volume will become smaller by a factor of 0.70., (d) Volume will become greater by a factor of 2.5., (Mains 2011), 15. The pressure exerted by 6.0 g of methane gas, in a 0.03 m3 vessel at 129°C is (Atomic masses:, C = 12.01, H = 1.01 and R = 8.314 J K–1 mol–1), (a) 215216 Pa, (b) 13409 Pa, (c) 41648 Pa, (d) 31684 Pa (Mains 2010), 16. Which of the following mixtures of gases does not, obey Dalton’s law of partial pressure?, (a) Cl2 and SO2, (b) CO2 and He, (c) O2 and CO2, (d) N2 and O2, (1996), 17. At what temperature, the rate of effusion of N2, would be 1.625 times than the rate of SO 2 at 50°C?, (a) 373°C, (b) 620°C, (c) 100°C, (d) 173°C, (1996), 18. 50 mL of hydrogen diffuses out through a small hole, of a vessel, in 20 minutes. The time taken by 40 mL, of oxygen to diffuse out is, (a) 32 minutes, (b) 64 minutes, (c) 8 minutes, (d) 12 minutes (1994), 19. Under what conditions will a pure sample of an, ideal gas not only exhibit a pressure of 1 atm but, also a concentration of 1 mole litre–1?, (R = 0.082 litre atm mol–1 deg–1), (a) At STP, (b) When V = 22.4 litres, (c) When T = 12 K, (d) Impossible under any conditions, (1993), , www.neetujee.com, , 20. The correct value of the gas constant ‘R’ is close to, (a) 0.082 litre-atmosphere K, (b) 0.082 litre-atmosphere K–1 mol–1, (c) 0.082 litre-atmosphere–1 K mol–1, (d) 0.082 litre–1 atmosphere–1 K mol., (1992), 21. Select one correct statement. In the gas equation,, PV = nRT, (a) n is the number of molecules of a gas, (b) V denotes volume of one mole of the gas, (c) n moles of the gas have a volume V, (d) P is the pressure of the gas when only one mole, of gas is present., (1992), 22. At constant temperature, in a given mass of an ideal gas, (a) the ratio of pressure and volume always remains, constant, (b) volume always remains constant, (c) pressure always remains constant, (d) the product of pressure and volume always, remains constant., (1991), 23. If P, V, M, T and R are pressure, volume, molar mass,, temperature and gas constant respectively, then for, an ideal gas, the density is given by, M, P, RT, (d) PM, (c), (b), (a), V, RT, PM, RT, (1989), 24. Correct gas equation is, P1V1 T1, V1T2 V2T1, , , (a), (b) P V T, P1, P2, 2 2, 2, V1V2, P1T1 P2V2,  PP, , (c), 1 2, (d) T T, V1, T2, , (1989), , 1 2, , 5.7, , Kinetic Energy and Molecular Speeds, , 25. By what factor does the average velocity of a gaseous, molecule increase when the temperature (in Kelvin), is doubled?, (a) 2.0, (b) 2.8, (c) 4.0, (d) 1.4, (2011), 26. The temperature of a gas is raised from 27°C to, 927°C. The root mean square speed of the gas, 927, (a) remains same, (b) gets, times, 27, (c) gets halved, (d) gets doubled. (1994), 27. The ratio among most probable velocity, mean, velocity and root mean square velocity is given by, (a) 1 : 2 : 3, (b) 1: 2 :, 3, (c), , 2 : 3:, , 8 / , , (d), , 2: 8/ : 3, (1993), , 28. The root mean square velocity at STP for the gases, H2, N2, O2 and HBr are in the order, , www.mediit.in

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States of Matter, (a) H2 < N2 < O2 < HBr (b) HBr < O2 < N2 < H2, (c) H2 < N2 = O2 < HBr (d) HBr < O2 < H2 < N2, (1991), 29. Root mean square velocity of a gas molecule is, proportional to, (a) m1/2, (b) m0, (c) m–1/2, (d) m, (1990), , 5.8, , Kinetic Molecular Theory of Gases, , 30. The energy absorbed by each molecule (A2) of, a substance is 4.4 × 10–19 J and bond energy per, molecule is 4.0 × 10–19 J. The kinetic energy of the, molecule per atom will be, (a) 2.2 × 10–19 J, (b) 2.0 × 10–19 J, (c) 4.0 × 10–20 J, (d) 2.0 × 10–20 J (2009), 31. If a gas expands at constant temperature, it indicates, that, (a) kinetic energy of molecules remains the same, (b) number of the molecules of gas increases, (c) kinetic energy of molecules decreases, (d) pressure of the gas increases., (2008), 32. Average molar kinetic energy of CO and N2 at same, temperature is, (a) KE1 = KE2, (b) KE1 > KE2, (c) KE1 < KE2, (d) can’t say anything. Both volumes are not given., (2000), , 39, , 37. Which is not true in case of an ideal gas?, (a) It cannot be converted into a liquid., (b) There is no interaction between the molecules., (c) All molecules of the gas move with same speed., (d) At a given temperature, PV is proportional to, the amount of the gas., (1992), , 5.9, , Behaviour of Real Gases - Deviation, from Ideal Gas Behaviour, , 38. A gas at 350 K and 15 bar has molar volume 20, percent smaller than that for an ideal gas under the, same conditions. The correct option about the gas, and its compressibility factor (Z) is, (a) Z < 1 and repulsive forces are dominant, (b) Z > 1 and attractive forces are dominant, (c) Z > 1 and repulsive forces are dominant, (d) Z < 1 and attractive forces are dominant., (NEET 2019), 39. A gas such as carbon monoxide would be most, likely to obey the ideal gas law at, (a) low temperatures and high pressures, (b) high temperatures and high pressures, (c) low temperatures and low pressures, (d) high temperatures and low pressures. (2015), 40. Maximum deviation from ideal gas is expected from, (a) CH4(g), (b) NH3(g), (c) H2(g), (d) N2(g), (NEET 2013), , 33. The average kinetic energy of an ideal gas, per, molecule in S.I. units, at 25°C will be, (a) 6.17 × 10–20 J, (b) 7.16 × 10–20 J, (c) 61.7 × 10–20 J, (d) 6.17 × 10–21 J (1996), , 41. For real gases van der Waals’ equation is written as,  an2 ,  p  2  (V  nb)  n RT where a and b are van,  V , , 34. At STP, 0.50 mol H2 gas and 1.0 mol He gas, (a) have equal average kinetic energies, (b) have equal molecular speeds, (c) occupy equal volumes, (d) have equal effusion rates., (1993), , der Waals’ constants. Two sets of gases are, (I) O2, CO2, H2 and He, (II) CH4, O2 and H2, The gases given in set-I in increasing order of b and, gases given in set-II in decreasing order of a, are, arranged below. Select the correct order from the, following :, (a) (I) He < H2 < CO2 < O2 (II) CH4 > H2 > O2, (b) (I) O2 < He < H2 < CO2 (II) H2 > O2 > CH4, (c) (I) H2 < He < O2 < CO2 (II) CH4 > O2 > H2, (d) (I) H2 < O2 < He < CO2 (II) O2 > CH4 > H2, (Mains 2012), 42. van der Waals’ real gas, acts as an ideal gas, at which, conditions?, (a) High temperature, low pressure, (b) Low temperature, high pressure, (c) High temperature, high pressure, (d) Low temperature, low pressure, (2002), , 35. Internal energy and pressure of a gas per unit, volume are related as, 3, 2, (b) P  E, (a) P  E, 2, 3, 1, (c) P  E, (d) P = 2E, (1993), 2, 36. A closed flask contains water in all its three states, solid, liquid and vapour at 0°C. In this situation, the, average kinetic energy of water molecules will be, (a) the greatest in all the three states, (b) the greatest in vapour state, (c) the greatest in the liquid state, (d) the greatest in the solid state., (1992), , www.neetujee.com, , www.mediit.in

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40, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 43. When is deviation more in the behaviour of a gas, from the ideal gas equation PV = nRT?, (a) At high temperature and low pressure, (b) At low temperature and high pressure, (c) At high temperature and high pressure, (d) At low temperature and low pressure, (1993), 44. A gas is said to behave like an ideal gas when the, relation PV/T = constant. When do you expect a real, gas to behave like an ideal gas?, (a) When the temperature is low., (b) When both the temperature and pressure are, low., (c) When both the temperature and pressure are, high., (d) When the temperature is high and pressure is, low., (1991), 45. In van der Waals’ equation of state for a non-ideal, gas, the term that accounts for intermolecular forces, is, (a) (V – b), (b) (RT)–1, a , , (c) P , (d) RT, (1990), , , V2 , , 1., (a), 11. (b), 21. (c), 31. (a), 41. (c), , 2., 12., 22., 32., 42., , (b), (b), (d), (a), (a), , 3., 13., 23., 33., 43., , (d), (a), (d), (d), (b), , 4., 14., 24., 34., 44., , (b), (a), (b), (a), (d), , 5., 15., 25., 35., 45., , 5.10 Liquefaction of Gases, 46. Given van der Waals’ constant for NH3, H2, O2 and, CO2 are respectively 4.17, 0.244, 1.36 and 3.59,, which one of the following gases is most easily, liquefied?, (a) NH3, (b) H2, (c) O2, (d) CO2 (NEET 2018), 47. An ideal gas, obeying kinetic theory of gases cannot, be liquefied, because, (a) it solidifies before becoming a liquid, (b) forces acting between its molecules are, negligible, (c) its critical temperature is above 0°C, (d) its molecules are relatively small in size. (1995), , 5.11 Liquid State, 48. The beans are cooked earlier in pressure cooker, because, (a) boiling point increases with increasing pressure, (b) boiling point decreases with increasing pressure, (c) extra pressure of pressure cooker softens the, beans, (d) internal energy is not lost while cooking in, pressure cooker., (2011), , ANSWER KEY, (a) 6., (c), (c) 16. (a), (d) 26. (d), (a) 36. (b), (c) 46. (a), , 7., 17., 27., 37., 47., , (d), (c), (d), (c), (b), , 8., (c), 18. (b), 28. (b), 38. (d), 48. (a), , 9., 19., 29., 39., , (c), (c), (c), (d), , 10., 20., 30., 40., , (None), (b), (d), (b), , Hints & Explanations, 1. (a) : HCl is polar (  0) and He is non-polar ( = 0), gives dipole-induced dipole interactions., 2. (b) : The strength of interaction follows the order :, van der Waals’ < hydrogen-bonding < dipole-dipole <, covalent. It is so because bond length of H-bond is larger, than that of a covalent bond., And also covalent bond is strongest because, the greater, the extent of overlapping, the stronger is the bond, formed., 3., , (d) : Mass of the gas = Mass of the cylinder, including gas – Mass of empty cylinder., So, mass of a gas can be determined by weighing the, , www.neetujee.com, , container in which it is enclosed., Thus, the statement (d) is wrong for gases., 4. (b) : V1 = 380 mL, P1 = 730 mm, V2 = ?,, P2 = 760 mm., From Boyle’s law, P1V1 = P2V2, P V 730  380, V 1 1,  365 mL, 2, 760, P2, 5. (a) : According to Charles’ law which states that, “The volume of the given mass of a gas increases or, 1, decreases by, of its volume at 0°C for each degree, 273, rise or fall of temperature at constant pressure.”, , www.mediit.in

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41, , States of Matter, V  V 1  t  at constant P and n., , t, 0  273 , 7, 6. (c) : Number of moles of N2  28  0.25 mol, 8, Number of moles of Ar  , 0.2 mol, 40, Mole fraction of N  0.25  0.25  0.55, 2, 0.25  0.2 0.45, Partial pressure of N2 gas = mole fraction × total pressure, = 0.55 × 27 = 14.85 ≈ 15 bar, m 1.8, 7. (d) : m = 1.8 g  n  ,  0.1 mol, M 18, T = 374°C = 647 K, P = 1 bar, R = 0.083 bar L K–1 mol–1, nRT 0.1  0.083  647, V, ,  5.37 L, P, 1, 8. (c) : Let the number of moles, 1 of each gas = x, Fraction of hydrogen escaped  x, , 11. (b) : According to Graham’s law of diffusion,,  r  M2,  r1 , 2, M1, d, M, Volume of gas diffused (V ), Rate of diffusion , Time taken (t), V1 / t1  M2, , , M1, V2 / t 2, r, , 1, , 1, , , , , , , , If same volume of two gases diffuse, then V1 = V2, t,  2  M2, t1, M1, Here t2 = 3t1, M1 = 4 u, M2 = ?, 3t, M2,  1,  3  M2, t, 4, 4, 1, M 2,  M2  36 u,  9 , 4, rA V / t A, ,  MB, , 12. (b) : We know that, MA, V / tB, rB, tB, MB,  MB  10, , MA, 20, 49, tA, 2,  10  MB, 100 MB,    , , , 20 , 49, 400 49, 49 100,  M,  12.25 u, B, 400, 13. (a) : pCO + pN2 = 1 atm, [ nCO = nN2], 2pN2 = 1, 1, p N   0.5 atm, 2, 2, T, 14. (a) : From ideal gas equation, V , P, Given T1 = 15 + 273 = 288 K, P1 = 1.5 bar, T2 = 25 + 273 = 298 K, P2 = 1 bar, 288, 298, V1 , i.e., V1 192 and V2 , 1.5, 1, V2 298, ,  1.55  1.6, V1 192, , , rO2  MH2 , , MO2, rH2, , nO2 / t, , , , , , nO / t, , , 2, , 1, , x, /t, 2, , 2, 1, 2  1 , 32, 16 4, , 1, ,   nO2  x, x/ t, 4, 8, 1, 2, Hence, fraction of oxygen escaped , 8, 9. (c) : PV = nRT, , Weight of the gas taken (W), w, PV  RT n , , Mol. mass of gas (M), M, , , w RT, , M V, , Density  Mass , d RT, P, , , M, Volume , PM 5  28, d, ,  3.41 g / mL, RT 0.0821  500, 10. (None) : According to Graham’s law of diffusion,, VB, r1, M2 , r  VA ,, d,  2, Br , A, r2, d1, M1, TA, TB, VA /TA, MB, , V /T, M, P, , B, , B, , A, , VA = VB, TA = 150 sec, TB = 200 sec, MB = 36, MA = ?, TB,  MB  200  36, TA, MA, 150, MA, 4,  36 or 4  4 36, , 3, 3  3 MA, MA, 36, or MA   3  3  20.25, 4 4, , www.neetujee.com, , , , , , 15. (c) : Given, mass of CH4, w = 6 g, Volume of CH4, V = 0.03 m3, T = 129°C = 129 + 273 = 402 K, R = 8.314 J K–1 mol–1, Molecular mass of CH4, M = 12.01 + 4 × 1.01 = 16.05, w, PV  nRT  RT, M, w RT 6 8.314  402, P,  , M V 16.05, 0.03, = 41647.7 Pa  41648 Pa, Sunlight, ,  SO2Cl2, 16. (a) : Cl2 + SO2  , (Sulphuryl chloride), , www.mediit.in

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42, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , Dalton’s law of partial pressure is applicable only in those, cases where gases are non-reacting. As Cl2 and SO2 reacts, to form SO2Cl2 so this law is not obeyed in given case., 17. (c) : r1 = 1.625r2 and T2 = 50°C = 323 K, r, We know that 1  M2  T1, M1 T2, r2, , 27. (d) : Most probable velocity, (ump ) , , 64 T1, , 28 323, 1.6252  28  323, , or 1.625 , , , , 8RT, M, Root mean square velocity, (u, , , , 21. (c) : In ideal gas equation, PV = nRT, n moles of the gas have volume V., 22. (d) : According to Boyle’s law at constant, 1, temperature, P  or PV = constant, V, 23. (d) : Ideal gas equation is, m, PV  nRT  RT, M, m, or PM  RT  dRT, [here d = density], V, PM,  d, RT, PV, PV P V, P1V1 T1, 24. (b) :, = constant or 1 1 2 2 , , P2V2 T2, T, T1, T2, , , , 8RT, M, , When T becomes 2T then, average velocity , i.e.,, , 8R(2T), M, , 2 or 1.41 times increase., , www.neetujee.com, , 2RT, M, , Mean velocity, (v ) , ,  373.15 K  100.15C, or T1 , 64, 18. (b) : Volume of hydrogen = 50 mL; Time for, diffusion (t) = 20 min and volume of oxygen = 40 mL., 50, = 2.5 mL/min, Rate of diffusion of hydrogen (r1 ) =, 20, 40, Rate of diffusion of oxygen (r2) =, mL/min, t, Since the molecular mass of hydrogen (M1) = 2 and that, of oxygen (M2) = 32, therefore, t, r1, 2.5  32   4  t  64 minutes,  M2 , , M1, 2, 16, r2, 40 t, n, 19. (c) : PV = nRT or P  RT  CRT, V, 1, Hence, 1 = 1 × 0.082 × T  T   12 K, 0.082, 20. (b), , 25. (d) : Average velocity , , 26. (d) : T1 = 27°C = 300 K and T2 = 927°C = 1200 K, We know that root mean square speed (v)  T ., Therefore root mean square speed of the gas, when its, 1200, T2, temperature is raised , ,  2 times, T1, 300, , r.m.s, ,  ump : v : ur.m.s, , ), , 3RT, , M, 2RT 8RT 3RT, , :, :, M, M, M, , 8, : 3, , 1, 1, 28. (b) : We know, PV  mnu2  Mu2, 3, 3, or u  3PV / M,  2:, , At STP, u  1, M, and molecular masses of H2, N2, O2 and HBr are 2, 28,, 32 and 81., 1, 29. (c) : PV  mNu2 ,, 3, here u = root mean square velocity., 3PV, 1, Now u2 , or u , mN, m, 30. (d) : Energy absorbed by each molecule, = 4.4 × 10–19 J, Energy required to break the bond = 4.0 × 10–19 J, Remaining energy to get converted to kinetic energy, = (4.4 × 10–19 – 4.0 × 10–19) J = 0.4 × 10–19 J per molecule,  Kinetic energy per atom = 0.2 × 10–19 J or 2 × 10–20 J, 31. (a) : The average translational K.E. of one molecule, of an ideal gas will be given by, K.E. 3/2 RT 3, Et , ,  kT, NA, 2, NA, where R/NA = Boltzmann constant i.e. Et  T, So, at constant temperature, K.E. of molecules remains, the same., 3, 32. (a) : K.E. = RT (for one mole of a gas), 2, As temperatures are same and KE is independent of, molecular mass, so KE1 = KE2., 33. (d) : Temperature (T) = 25°C = 298 K., Therefore, K.E. per molecule, 3RT 3  8.314  298, , ,  6.17  1021 J, 2NA 2  (6.02 1023 ), , www.mediit.in

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States of Matter, , 43, , 34. (a) : Because average kinetic energy depends only, 3, on temperature K.E.  kT, 2 1, 1, 35. (a) : PV  mnu2  Mu2, 3, 2 1, , 2, , 2, , 3, , 1, , , Mu2  E , ,   Mu  E,  2, , , , 3 2, 3, 2, or P  E per unit volume., 3, 36. (b) : Velocity and hence average K.E. of water, molecules is maximum in the gaseous state., 37. (c) : Molecules in an ideal gas move with different, speeds. Due to collision between the particles their speed, changes., 38. (d) : Videal = V, Vreal = V – 0.2 V = 0.8 V, V, Z  V real  0.8, ideal, if value of Z < 1 then attractive forces are dominant., 39. (d) : Real gases show ideal gas behaviour at high, temperatures and low pressures., 40. (b) : NH3 is a polar molecule, thus more attractive, forces between NH3 molecules., 41. (c) : van der Waals’ gas constant ‘a’ represent, intermolecular forces of attraction of gaseous molecules, and van der Waals’ gas constant ‘b’ represent effective, size of molecules. Therefore order should be, (I) H2 < He < O2 < CO2 (II) CH4 > O2 > H2, 42. (a) : At low pressure and high temperature van, der Waals real gas acts as ideal gas and observed to, obey PV = nRT relation. At very low pressure when, the gas-volume is quite large the space occupied by the, molecules themselves becomes negligible comparatively, and because the molecules are then far apart, the force of, mutual attraction becomes too feeble, the real gas would, satisfy the postulates of kinetic theory. As temperature, , is raised, the volume of the gas increases and we can, consider , n2a  term as P and at low pressure, P , 2 , , (V – nb  VV., ), term, as, , n2a , P, (V  nb)  nRT (van der Waals’ equation), , , , V 2 , This equation becomes PV = nRT, This is an ideal gas equation., , , , , 43. (b) : At low temperature and high pressure, there is, a deviation from the ideal behaviour in gases., 44. (d) : At high temperature and low pressure the effect, of a/V2 and b is negligible., As we know, PV = nRT (Ideal gas equation), PV, PV = RT or, 1, RT,  Z =1, [Z is compressibility factor], Hence gas shows ideal behaviour., 45. (c) : van der Waals’ equation for 1 mole is, a, , P  (V  b)  RT, , , V2  , a , , Here, P  2  represents the intermolecular forces,  V , and (V – b) is the correct volume., 46. (a) : van der Waals’ constant ‘a’ signifies the, intermolecular forces of attraction between the particle, of gas. So, higher the value of ‘a’, easier will be the, liquefaction of gas., 47. (b) : A gas can only be liquefied, if some forces, of attraction are acting in its molecules. According to, kinetic theory, an ideal gas is devoid of force of attraction, in its molecules, therefore it cannot be liquefied., 48. (a) : More is the pressure, greater will be the boiling, point., , , , , , , , , , , , , , , , , , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 6, , 6.1, 1., , 2., , 3., , 5., , 6., , Which of the following options is not correct ?, (a) Sisothermal > Sadiabatic, (b) T = T, A, B, (c) W isothermal > W adiabatic, , Thermodynamic Terms, , Which of the following are not state functions?, (I) q + w, (II) q, (III) w, (IV) H – TS, (a) (I), (II) and (III), (b) (II) and (III), (c) (I) and (IV), (d) (II), (III) and (IV), (2008), In a closed insulated container a liquid is stirred, with a paddle to increase the temperature, which of, the following is true?, (a) E = W  0, q = 0 (b) E = W = q  0, (c) E = 0, W = q  0 (d) W = 0, E = q  0, (2002), , 7., , A gas is allowed to expand in a well insulated, container against a constant external pressure of, 2.5 atm from an initial volume of 2.50 L to a final, volume of 4.50 L. The change in internal energy U, of the gas in joules will be, (a) –500 J, (b) –505 J, (c) +505 J, (d) 1136.25 J, (NEET 2017), , 9., , Equal volumes of two monatomic gases, A and B at, same temperature and pressure are mixed. The ratio, of specific heats (CP/CV) of the mixture will be, (a) 0.83, (b) 1.50, (c) 3.3, (d) 1.67, (2012), , Applications, , The correct option for free expansion of an ideal gas, under adiabatic condition is, (a) q = 0, T = 0 and w = 0, (b) q = 0, T < 0 and w > 0, (c) q < 0, T = 0 and w = 0, (d) q > 0, T > 0 and w > 0, (NEET 2020), Under isothermal conditions, a gas at 300 K expands, from 0.1 L to 0.25 L against a constant external, pressure of 2 bar. The work done by the gas is, [Given that 1 L bar = 100 J], (a) 30 J, (b) –30 J, (c) 5 kJ, (d) 25 J (NEET 2019), Reversible expansion of an, ideal gas under isothermal, and adiabatic conditions are, as shown in the figure., AB  Isothermal expansion, AC  Adiabatic expansion, , www.neetujee.com, , (d) T > T, (Odisha NEET 2019), c, A, An ideal gas expands isothermally from 10–3 m3, to 10–2 m3 at 300 K against a constant pressure of, 105 N m–2. The work done on the gas is, (a) +270 kJ, (b) –900 J, (c) +900 kJ, (d) –900 kJ, (Odisha NEET 2019), , 8., , Which of the following is the correct equation?, (a) U = W + Q, (b) U = Q – W, (c) W = U + Q, (d) None of these (1996), , 6.2, 4., , Thermodynamics, , 10. Which of the following is correct option for free, expansion of an ideal gas under adiabatic condition?, (a) q = 0, T  0, w = 0 (b) q  0, T = 0, w = 0, (c) q = 0, T = 0, w = 0 (d) q = 0, T < 0, w  0, (2011), 11. Three moles of an ideal gas expanded spontaneously, into vacuum. The work done will be, (a) infinite, (b) 3 Joules, (c) 9 Joules, (d) zero., (Mains 2010), 12. Assume each reaction is carried out in an open, container. For which reaction will H = E?, (a) 2CO(g) + O2(g)  2CO2(g), (b) H2(g) + Br2(g)  2HBr(g), , www.mediit.in

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Thermodynamics, , 45, , (c) C(s) + 2H2O(g)  2H2(g) + CO2(g), (d) PCl5(g)  PCl3(g) + Cl2(g), (2006), 13. The work done during the expansion of a gas from a, volume of 4 dm3 to 6 dm3 against a constant external, pressure of 3 atm is (1 L atm = 101.32 J), (a) – 6 J, (b) – 608 J, (c) + 304 J, (d) – 304 J, (2004), 14. For the reaction,, C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l), at constant temperature, H – E is, (a) + RT, (b) –3RT, (c) +3RT, (d) –RT, , (2003), , 15. The molar heat capacity of water at constant, pressure, C , is 75 J K–1 mol–1. When 1.0 kJ of heat is, supplied to 100 g of water which is free to expand,, the increase in temperature of water is, (b) 2.4 K, (a) 1.2 K, (c) 4.8 K, (d) 6.6 K, (2003), 16. When 1 mol of gas is heated at constant volume, temperature is raised from 298 to 308 K. Heat, supplied to the gas is 500 J. Then which statement is, correct?, (a) q = w = 500 J, E = 0, (b) q = E = 500 J, w = 0, (c) q = w = 500 J, E = 0, (d) E = 0, q = w = –500 J, (2001), 17. For the reaction,, C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l), which one is true?, (a) H = E – RT, (b) H = E + RT, (c) H = E + 2RT, (d) H = E – 2RT (2000), 18. In an endothermic reaction, the value of H is, (a) negative, (b) positive, (c) zero, (d) constant., (1999), 19. One mole of an ideal gas at 300 K is expanded, isothermally from an initial volume of 1 litre to, 10 litres. The E for this process is, (R = 2 cal mol–1 K–1), (a) 1381.1 cal, (b) zero, (c) 163.7 cal, (d) 9 L atm, (1998), 20. During isothermal expansion of an ideal gas, its, (a) internal energy increases, (b) enthalpy decreases, (c) enthalpy remains unaffected, (d) enthalpy reduces to zero., (1994, 1991), 21. For the reactionm, N2 + 3H2  2NH 3, H = ?, (a) E + 2RT, (b) E – 2RT, (c) H = RT, (d) E – RT, (1991), , www.neetujee.com, , 22. If H is the change in enthalpy and E, the change, in internal energy accompanying a gaseous reaction,, then, (a) H is always greater than E, (b) H < E only if the number of moles of the, products is greater than the number of moles of, the reactants, (c) H is always less than E, (d) H < E only if the number of moles of products, is less than the number of moles of the reactants., (1990), , 6.4, , Enthalpy Change, rH of a Reaction Reaction Enthalpy, , 23. Three thermochemical equations are given below :, + O  CO ;  H° = x kJ mol–1, (i) C, (graphite), , 2(g), , 2(g), , r, , 1, (ii) C(graphite) + O2(g)  CO(g) ; rH° = y kJ mol–1, 2, 1, (iii) CO(g) + O2(g)  CO2(g); rH° = z kJ mol–1, 2, Based on the above equations, find out which of the, relationship given below is correct., (a) z = x + y, (b) x = y + z, (c) y = 2z – x, (d) x = y – z, (Karnataka NEET 2013), 24. Standard enthalpy of vaporisation vapH° for water, at 100°C is 40.66 kJ mol–1. The internal energy of, vaporisation of water at 100°C (in kJ mol–1) is, (a) +37.56, (b) – 43.76, (c) + 43.76, (d) + 40.66, (Assume water vapour to behave like an ideal gas), (2012), 25. Consider the following processes :, H (kJ/mol), 1/2A  B, +150, 3B  2C + D, 125, E + A  2D, +350, For B + D  E + 2C, H will be, (a) 525 kJ/mol, (b) –175 kJ/mol, (c) –325 kJ/mol, (d) 325 kJ/mol, (Mains 2011), 26. The following two reactions are known, Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g); H = –26.8 kJ, FeO(s) + CO(g)  Fe(s) + CO2(g); H = – 16.5 kJ, The value of H for the following reaction, Fe2O3(s) + CO(g)  2FeO(s) + CO2(g) is, (a) + 10.3 kJ, (b) – 43.3 kJ, (c) – 10.3 kJ, (d) + 6.2 kJ (Mains 2010), , www.mediit.in

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Thermodynamics, , 47, , 41. Given that bond energies of H – H and Cl – Cl are, 430 kJ mol–1 and 240 kJ mol–1 respectively and Hf, for HCl is –90 kJ mol–1, bond enthalpy of HCl is, (a) 380 kJ mol–1, (b) 425 kJ mol–1, –1, (c) 245 kJ mol, (d) 290 kJ mol–1 (2007), 42. The absolute enthalpy of neutralisation of the, reaction :, MgO(s) + 2HCl(aq)  MgCl2(aq) + H2O(l) will be, (a) –57.33 kJ mol–1, (b) greater than –57.33 kJ mol–1, (c) less than –57.33 kJ mol–1, (d) 57.33 kJ mol–1, (2005), 43. If the bond energies of H–H, Br–Br, and H–Br are, 433, 192 and 364 kJ mol–1 respectively, the H° for, the reaction H2(g) + Br2(g)  2HBr(g) is, (a) –261 kJ, (b) +103 kJ, (d) –103 kJ, , (c) +261 kJ, , 6.6, , (2004), , Spontaneity, , 44. For the reaction, 2Cl(g)  Cl2(g), the correct option is, (a) rH > 0 and rS > 0 (b) rH > 0 and rS < 0, (c) rH < 0 and rS > 0 (d) rH < 0 and rS < 0, (NEET 2020), 45. In which case change in entropy is negative?, (a) 2H(g)  H2(g), (b) Evaporation of water, (c) Expansion of a gas at constant temperature, (NEET 2019), (d) Sublimation of solid to gas, 46. For a given reaction, H = 35.5 kJ mol–1 and, S = 83.6 J K–1 mol–1. The reaction is spontaneous, at (Assume that H and S do not vary with, temperature.), (a) T > 425 K, (b) all temperatures, (c) T > 298 K, (d) T < 425 K, (NEET 2017), 47. For a sample of perfect gas when its pressure is, changed isothermally from pi to pf , the entropy, change is given by, p, p , (b) S  nR ln i , (a) S  nR ln f , p , p ,  i ,  f , p,  f ,  pi , , , , , , , (c) S  nRT ln  , pi , , , , (d) S  RT ln  , p f , (NEET-II 2016), 48. The correct thermodynamic conditions for the, spontaneous reaction at all temperatures is, (a) H < 0 and S > 0 (b) H < 0 and S < 0, (c) H < 0 and S = 0 (d) H > 0 and S < 0, (NEET-I 2016), , , www.neetujee.com, , 49. Consider the following liquid-vapour equilibrium., Liquid, Vapour, Which of the following relations is correct?, d ln P Hv, d ln P H, , v, (b), (a), 2  2, RT2, dT, dT, T, (d) d ln P Hv, (c) d lnG Hv, , , dT, RT, dT2, RT2, (NEET-I 2016), 50. Which of the following statements is correct for the, spontaneous adsorption of a gas?, (a) S is negative and, therefore H should be, highly positive., (b) S is negative and therefore, H should be, highly negative., (c) S is positive and therefore, H should be, negative., (d) S is positive and therefore, H should also be, highly positive., (2014), 2XO2(g), 51. For the reaction, X2O4(l), –1, , U = 2.1 kcal, S = 20 cal K at 300 K, Hence, G is, (a) 2.7 kcal, (b) – 2.7 kcal, (c) 9.3 kcal, (d) – 9.3 kcal, , (2014), , 52. A reaction having equal energies of activation for, forward and reverse reactions has, (a) H = 0, (b) H = G = S = 0, (c) S = 0, (d) G = 0 (NEET 2013), 53. In which of the following reactions, standard, reaction entropy change (S°) is positive and, standard Gibbs energy change (G°) decreases, sharply with increasing temperature?, 1, (a) C, + O  CO, (graphite), , 2(g), , (g), , 1 2, (b) CO(g) + O2(g)  CO2(g), 2, 1, (c) Mg + O  MgO, (s), (s), 2 2(g)1, 1, 1, (d) C(graphite) + O2(g)  CO2(g), , (2012), , 2, 2, 2, 54. The enthalpy of fusion of water is 1.435 kcal/mol., The molar entropy change for the melting of ice at, 0°C is, (a) 10.52 cal/(mol K) (b) 21.04 cal/(mol K), (c) 5.260 cal/(mol K) (d) 0.526 cal/(mol K), (2012), 55. If the enthalpy change for the transition of liquid, water to steam is 30 kJ mol–1 at 27°C, the entropy, change for the process would be, , www.mediit.in

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48, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, –1, , –1, , –1, , (a) 10 J mol K, (c) 0.1 J mol–1 K–1, , –1, , (b) 1.0 J mol K, (d) 100 J mol–1 K–1, (2011), , 56. Standard entropies of X2, Y2 and XY3 are 60, 40, and 50 J K–1 mol–1 respectively. For the reaction, 1/2X2 + 3/2Y2  XY3, H = –30 kJ, to be at, equilibrium, the temperature should be, (a) 750 K, (b) 1000 K, (c) 1250 K, (d) 500 K, (2010), 57. For vaporization of water at 1 atmospheric, pressure, the values of H and S are, 40.63 kJ mol–1 and 108.8 J K–1 mol–1, respectively., The temperature when Gibbs’ energy change (G), for this transformation will be zero, is, (a) 273.4 K, (b) 393.4 K, (c) 373.4 K, (d) 293.4 K (Mains 2010), 58. The values of H and S for the reaction,, C(graphite) + CO2(g)  2CO(g), are 170 kJ and 170 J K–1, respectively. This reaction, will be spontaneous at, (a) 910 K, (b) 1110 K, (c) 510 K, (d) 710 K, (2009), 59. For the gas phase reaction,, PCl5(g)  PCl3(g) + Cl2(g), which of the following conditions are correct?, (a) H < 0 and S < 0 (b) H > 0 and S < 0, (c) H = 0 and S < 0 (d) H > 0 and S > 0, (2008), 60. Identify the correct statement for change of, Gibbs’ energy for a system (Gsystem) at constant, temperature and pressure., (a) If G < 0, the process is not spontaneous., system, , (b), (c) If, If Gsystem > 0, the process is spontaneous., Gsystem = 0, the system has attained, equilibrium., (d) If Gsystem = 0, the system is still moving in a, particular direction., (2006), 61. The enthalpy and entropy change for the reaction:, Br2(l) + Cl2(g)  2BrCl, –1, , (g), , –1, , (2005), , 63. A reaction occurs spontaneously if, (a) TS < H and both H and S are +ve, (b) TS > H and H is +ve and S is –ve, (c) TS > H and both H and S are +ve, (d) TS = H and both H and S are +ve (2005), 64. Standard enthalpy and standard entropy changes, for the oxidation of ammonia at 298 K are, –382.64 kJ mol–1 and –145.6 J mol–1, respectively., Standard Gibbs’ energy change for the same reaction, at 298 K is, (a) – 221.1 kJ mol–1, (b) – 339.3 kJ mol–1, (c) – 439.3 kJ mol–1, (d) – 523.2 kJ mol–1, (2004), 65. Considering entropy (S) as a thermodynamic, parameter, the criterion for the spontaneity of any, process is, (a) Ssystem + Ssurroundings > 0, (b) Ssystem – Ssurroundings > 0, (c) Ssystem > 0 only, (d) Ssurroundings > 0 only., , (2004), –1, , –1, , 66. What is the entropy change (in J K mol ) when, one mole of ice is converted into water at 0°C? (The, enthalpy change for the conversion of ice to liquid, water is 6.0 kJ mol–1 at 0°C.), (a) 20.13, (b) 2.013, (c) 2.198, (d) 21.98, (2003), 67. The densities of graphite and diamond at 298 K are, 2.25 and 3.31 g cm–3, respectively. If the standard, free energy difference (G°) is equal to 1895 J mol–1,, the pressure at which graphite will be transformed, into diamond at 298 K is, (a) 9.92 × 108 Pa, (b) 9.92 × 107 Pa, (c) 9.92 × 106 Pa, (d) 9.92 × 105 Pa (2003), 68. Unit of entropy is, (a) J K–1 mol–1, (c) J–1K–1 mol–1, , (b) J mol–1, (d) J K mol–1, , (2002), , –1, , are 30 kJ mol and 105 J K mol respectively., The temperature at which the reaction will be in, equilibrium is, (a) 300 K, (b) 285.7 K, (c) 273 K, (d) 450 K, (2006), 62. Which of the following pairs of a chemical reaction, is certain to result in a spontaneous reaction?, (a) Exothermic and increasing disorder, , www.neetujee.com, , (b) Exothermic and decreasing disorder, (c) Endothermic and increasing disorder, (d) Endothermic and decreasing disorder, , 69. 2 moles of ideal gas at 27°C temperature is expanded, reversibly from 2 lit. to 20 lit. Find entropy change., (R = 2 cal/mol K), (a) 92.1, (b) 0, (c) 4, (d) 9.2, (2002), 70. PbO2  PbO; G298 < 0, SnO2  SnO; G298 > 0, Most probable oxidation state of Pb and Sn will be, , www.mediit.in

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Thermodynamics, , 49, , (a) Pb4+, Sn4+, 2+, , (b) Pb4+, Sn2+, , 2+, , 2+, , (c) Pb , Sn, , 4+, , (c) At absolute zero of temperature, entropy of a, perfectly crystalline substance is taken to be, zero., (d) At 0°C, the entropy of a perfectly crystalline, substance is taken to be zero., (1998), , (2001), , (d) Pb , Sn, , 71. Cell reaction is spontaneous when, (a) G° is negative, (b) G° is positive, (c) E°red is positive, (d) E°red is negative., , 6.7, , (2000), 72. Identify the correct statement regarding entropy., (a) At absolute zero of temperature, the entropy of, all crystalline substances is taken to be zero., (b) At absolute zero of temperature, the entropy of a, perfectly crystalline substance is +ve., , Gibbs Energy Change and Equilibrium, , 73. Following reaction occurring in an automobile, 2C8H18(g) + 25O2(g)  16CO2(g) + 18H2O(g), The sign of H, S and G would be, (a) –, +, +, (b) +, +, –, (c) +, –, +, (d) –, +, –, (1994), , ANSWER KEY, , 1., 11., 21., 31., 40., 50., 60., 70., , (b), (d), (b), (d), (d), (b), (c), (d), , 2., 12., 22., 32., 41., 51., 61., 71., , (a), (b), (d), (a), (b), (b), (b), (a), , 3., 13., 23., 33., 42., 52., 62., 72., , (b), (b), (b), (d), (c), (a), (a), (c), , 4., 14., 24., 34., 43., 53., 63., 73., , (a), (b), (a), (c), (d), (a), (c), (d), , 5., 15., 25., 35., 44., 54., 64., , (b) 6., (b) 16., (b) 26., (None), (d) 45., (c) 55., (b) 65., , (d), (b), (d), (a), (d), (a), , 7., 17., 27., 36., 46., 56., 66., , (b), (a), (c), (b), (a), (a), (d), , 8., 18., 28., 37., 47., 57., 67., , (b), (b), (a), (c), (b), (c), (a), , 9., 19., 29., 38., 48., 58., 68., , (d), (b), (a), (b), (a,c), (b), (a), , 10., 20., 30., 39., 49., 59., 69., , (c), (c), (a), (a), (b), (d), (d), , Hints & Explanations, 1. (b) : State functions or state variables are those, which depend only on the state of the system and not on, how the state was reached., q  w  E (internal energy),  State functions, H  TS  G (free energy) , , Path function depends on the path followed during a, process. Work and heat are the path functions., 2. (a) : The mathematical form of first law of, thermodynamics : q = E + W, Since the system is closed and insulated, q = 0, Paddle work is done on system.  W  0., Temperature and hence internal energy of the system, increases.  E  0., 3. (b) : This is the mathematical relation of first law of, thermodynamics. Here U = change in internal energy;, Q = heat absorbed by the system and W = work done by, the system., 4. (a) : For free expansion of an ideal gas, Pex = 0,, w = –Pex V = 0, For adiabatic process, q = 0, According to first law of thermodynamics,, U = q + w = 0, , www.neetujee.com, , As internal energy of an ideal gas is a function of, temperature, U = 0,  T = 0, 5. (b) : Expansion of a gas against a constant external, pressure is an irreversible process. The work done in an, irreversible process, = – Pext V = – Pext (V2 – V1) = –2 (0.25 – 0.1), = – 2 × 0.15 L bar = – 0.30 × 100 J = – 30 J, 6. (d) : For an ideal gas, internal energy is a function, of temperature. Final temperature i.e., TC for adiabatic, process is less than its initial temperature i.e., TA,  TC < TA, 7. (b) : w = –PdV = –P(V2 – V1), = –105 N m–2 (10–2 – 10–3) m3 = –105 N m–2 (9 × 10–3) m3, = –9 × 102 N m = –900 J, (Q 1 J = 1 N m), 8. (b) : w = –PextV = –2.5(4.50 – 2.50), = – 5 L atm = – 5 × 101.325 J = – 506.625 J, U = q + w, As, the container is insulated, thus q = 0, Hence, U = w = – 506.625 J, 9. (d) : C, gas mixture of same, 5 P for monoatomic, 3, volume  R, C  R, 2, , V, , 2, , www.mediit.in

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50, , 5, C, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , R, , 5, ,  2   1.67, CV 3 R 3, 2, 10. (c) : For free expansion of an ideal gas under, adiabatic condition q = 0, T = 0, w = 0., For free expansion, w = 0, adiabatic process, q = 0, U = q + w = 0, Internal energy remain constant means T = 0., 11. (d) : Since the ideal gas expands spontaneously into, vacuum, Pext = 0, hence work done is also zero., , , P, , 12. (b) : H = E + ngRT, For H + Br  2HBr, 2(g), (g), n 2(g), g = 2 – (1 + 1) = 0. i.e. H = E, 13. (b) : Work = –Pext × volume change, = –3 × (6 – 4) × 101.32 = 6 × 101.32, = – 607.92 J  – 608 J, 14. (b) : C H + 5O  3CO + 4H O, 2(g), 2(g), 2 (l), 3 8(g), n, g = 3 – 6 = –3, H = E + PV or H – E = PV, H – E = ngRT = –3RT, 15. (b) : Molar heat capacity = 75 J K–1 mol–1, 18 g of water = 1 mole = 75 J K–1 mol–1, 75, 1 g of water = J K–1, 18, 75, 100 g of water   100 J K1, 75, Q = m · C · T 18 1000  100   T, or, 18, 10 18,  T ,  2.4 K, 75, 16. (b) : H = E + PV, When V = 0; w = 0., H = E + 0 or H = E, As E = q + w, E = q, In the present problem, H = 500 J,, H = E = 500 J, q = 500 J, w = 0, 17. (a) : H = E + PV, also PV = nRT (ideal gas equation), or PV = ngRT, ng = Change in number of gaseous moles,  H = E + ngRT  ng = 2 – 3 = –1,  H = E – RT, 18. (b) : In endothermic reactions, energy of reactants, is less than energy of products. Thus, ER < EP., H = EP – ER = +ve, 19. (b) : Change in internal energy depends upon, temperature. At constant temperature, the internal, energy of the gas remains constant, so E = 0., , www.neetujee.com, , 20. (c) : During isothermal expansion of an ideal gas,, T = 0, E = 0  H = 0, H = E + PV,  H = E + PV) = E + (nRT),  H = E + nRT = 0 + 0 = 0, Change in enthalpy is zero, means its enthalpy remains, same or unaffected., 21. (b) : ng = 2 – 4 = –2, H = E – 2RT, 22. (d) : If np < nr; ng = np – nr = –ve., Hence, H < E., 23. (b) : According to Hess’s law, equation (i) is equal to, equations (ii) + (iii) i.e., x = y + z, 24. (a) : vapH° = 40.66 kJ mol–1, T = 100 + 273 = 373 K, E = ?, H = E + n RT  E = H – n RT, g, g, n, g = number of gaseous moles of products, – number of gaseous moles of reactants, H2O(l) H2O(g ), ng = 1 – 0 = 1, E = H – RT, E = (40.66 × 103) – (8.314 × 373), = 37559 J/mol or 37.56 kJ/mol, 25. (b) : Adding all the equations, we get, H, A  2B, 300 kJ/mol, 3B  2C + D, –125 kJ/mol, 2D  A + E, –350 kJ/mol, B + D  E + 2C ; H = (300 – 125 – 350), = –175 kJ/mol, 26. (d) : Fe O + 3CO  2Fe + 3CO ;, 2 3(s), , (g), , (s), , 2(g), , H = – 26.8 kJ ...(i), FeO(s) + CO(g)  Fe(s) + CO2(g) ; H = – 16.5 kJ, ...(ii), Fe2O3(s) + CO(g)  2FeO(s) + CO2(g) ; H = ?, ..(iii), Eq. (iii) can be obtained as :, (i) – 2(ii) = –26.8 – 2(–16.5) = –26.8 + 33.0 = +6.2 kJ, 27. (c) : For (c), H°reaction, = H°f (XeF4) – [H°f (Xe) + 2H°f (F2)], Enthalpies of formation of elementary substances Xe and, F2 are taken as zero., Thus, H°reaction = H°f (XeF4), 28. (a) : (i) C(s) + O2(g)  CO2(g); Hi = –94 kcal/mole, (ii) 2H2(g)+O2(g)  2H2O(l); Hii = –68 × 2 kcal/mole, (iii) CH4(g) + 2O2(g)  CO2(g) + 2H2O(l);, Hiii = –213 kcal/mole, (iv) C(s) + 2H2(g)  CH4(g); Hiv = ?, By applying Hess’s law, we can compute Hiv.,  Hiv = Hi + Hii – Hiii, = (–94 – 68 × 2 + 213) kcal = –17 kcal, , www.mediit.in

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Thermodynamics, , 51, , 29. (a) : H°f = H°f (products) – H°f (reactants), For the given reaction,, 2H2O2(l)  2H2O(l) + O2(g), H°f = 2 × H°f (H, , 2 O), , – 2 × H°f (H, –1, , 2 O2 ), , = 2 × –286 kJ mol – 2 × (–188) kJ mol–1, = –196 kJ mol–1, 30. (a) : CH + 2O  CO + 2H O, H = –x, ...(i), 2, 2, 2, 1, 34, CH3OH + O2  CO2 + 2H2O, H2 = –y, ...(ii), 2, Subtracting (ii) from (i), we get, 1, CH4 + O2  CH3OH, H3 = –ve, 2, i.e., –x – (–y) = –ve, y – x = – ve. Hence, x > y., 3, 31. (d) : S + O2  SO3 + 2x kcal, ...(i), 2, 1, SO2 + O2  SO3 + y kcal ................................. (ii), 2, By subtracting equation (ii) from (i) we get,, S + O2  SO2 + (2x – y) kcal, The heat of formation of SO2 is (2x – y) kcal/mole., 32. (a) : C(s) + O2(g)  CO2(g); H° = – x kJ .............. (i), 1, , CO(g) + 2 O2(g)  CO2(g); H° = 2y kJ ..................... (ii), By subtracting equation (ii) from (i) we get,, 1, C + O  CO ;, (s), 2 2(g)  y  (g)y  2x, H   x  , , kJ, , ,  2 , 2, 33. (d) : C2H4 + 3O2  2 CO2 + 2H2O, H° = H°products – H°reactants, = 2 × (– 394) + 2 × (– 286) – (52 + 0) = – 1412 kJ/mol, 34. (c) : Let B.E. of X2, Y2 and XY are x kJ mol–1,, 0.5x kJ mol–1 and x kJ mol–1 respectively., 1, 1, X2  Y2  XY ; H  200 kJ mol1, 2, 2, H = (B.E.),  1 Reactants –1(B.E.)Products, ,   200   (x)   (0.5x) [1  (x)], 2, , 2, B.E. of X2 = x = 800 kJ mol–1, , 35. (None) : Given :, C(s) + O2(g), CO2(g), H = – 393.5 kJ/mol,  Amount of heat released on formation of, 44 g CO2 = 393.5 kJ,  Amount of heat released on formation of, 393.5,  35.2 = 314.8  315 kJ, 35.2 g of CO2 , 44, Note : –ve or +ve sign considering the reaction is, exothermic or endothermic., , www.neetujee.com, , 36. (b) : CH4 + 2O2  CO2 + 2H2O, C3H8 + 5O2  3CO2 + 4H2O, No. of moles in gaseous mixture, 5, CH + C H =  0.22 moles, 4, 38, 22.4, 16, No. of moles of O =, = 0.71 moles, 2, 22.4, Let x moles of CH4 is there in a gaseous mixture so,, number of moles of C3H8 would be 0.22 – x. Then moles, of O2 consumed,, 2x + (0.22 – x)5 = 0.71 or x = 0.13, Total amount of heat liberated, = 0.13 × 890 + 0.09 × 2220 = 315.5 kJ, 37. (c) : The dissociation energy of H – H bond is, 869.6,  434.8 kJ, 2, 38. (b) : For the given reaction, enthalpy of reaction can, be calculated as, = B.E.(reactants) – B.E.(products), = [B.E.(C=C) + B.E.(H–H) + 4 × B.E.(C–H)], – [B.E.(C–C) + 6 × B.E.(C–H)], = [606.10 + 431.37 + 4 × 410.50]–[336.49 + 6 × 410.50], = 2679.47 – 2799.49 = –120.02 kJ mol–1, 39. (a) : H2 + Cl2  2HCl, Hreaction = (B.E)reactants – (B.E)products, = [(B.E)H – H + (B.E)Cl – Cl] – [2B.E(H – Cl)], = 434 + 242 – (431) × 2, Hreaction = –186 kJ, Heat of formation is the amount of heat absorbed or, evolved when one mole of substance is directly obtained, from its constituent elements., 186, –1, Hence, enthalpy of formation of HCl  2 = –93 kJ mol, 40. (d) : The amount of heat absorbed or released, when 1 mole of a substance is directly obtained from its, constituent elements is called the heat of formation or, enthalpy of formation., Equation (i) represents neutralisation reaction, (iii), represents hydrogenation reaction and (iv) represents, combustion reaction., –1, Thus, enthalpy of formation of H2O(l) is –X2 kJ mol ., 1, 41. (b) : H  1 Cl  HCl, 2 2, 2 2, H =  B.E.(reactants) –  B.E.(products), 1, = [B.E.(H ) + B.E.(Cl )] – B.E.(HCl) = –90, 2, 2, 2, 1, (430  240)  B.E.(HCl)  90, 2, 1, B.E.(HCl)  (430  240)  90  425 kJ mol1, 2, , www.mediit.in

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52, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 42. (c) : MgO is the oxide of weak base and we know, that heat of neutralisation of 1 eq. of strong acid with, strong base is –57.33 kJ/mol.,  With weak base some heat is absorbed in dissociation, of weak base.,  Heat of neutralisation of weak base with strong acid, will be less than –57.33 kJ/mol., 43. (d) : H – H + Br – Br  2H – Br, 433 + 192, 2 × 364, = 625, = 728, Energy absorbed, , Energy released, , Net energy released = 728 – 625 = 103 kJ, i.e. H° = – 103 kJ, 44. (d) : In the reaction, 2Cl(g) Cl2(g), the randomness, decreases as 2 moles of Cl(g) are converted to 1 mole of, Cl2(g), thus, rS < 0., And this is an exothermic reaction, thus, rH < 0., 45. (a) : If ng < 0 then S < 0, 46. (a) : For a spontaneous reaction,, G < 0 i.e., H – TS < 0, H, T, 35.5 1000, , S,  424.6  425 K, T, ,  83.6,  T > 425 K, 47. (b) : For an ideal gas undergoing reversible, expansion, when temperature changes from Ti to Tf and, pressure changes, pf ,, Tf from ppi to, i, ln, , nR, ln, S  nC, p, Ti, pf, For an isothermal process, T = T so, ln 1 = 0, i, , f, , p,  S  nR ln i, pf, 48. (a, c) : G = H – TS, If H < 0 and S > 0, G = (–ve) – T(+ve), then at all temperatures, G = –ve, spontaneous reaction., If H < 0 and S = 0, G = (–ve) – T(0) = –ve at all temperatures., 49. (b) : This is Clausius—Clapeyron equation., 50. (b) : Using Gibbs’-Helmholtz equation,, G = H – TS, During adsorption of a gas, entropy decreases i.e. S < 0, For spontaneous adsorption, G should be negative,, which is possible when H is highly negative., 51. (b) : H = U + ngRT, Given, U = 2.1 kcal, ng = 2,, R = 2 × 10–3 kcal, T = 300 K, , www.neetujee.com, ,  H = 2.1 + 2 × 2 × 10–3 × 300 = 3.3 kcal, Again, G = H – TS, Given, S = 20 × 10–3 kcal K–1, On putting the values of H and S in the equation, we, get G = 3.3 – 300 × 20 × 10–3, = 3.3 – 6 × 103 × 10–3 = – 2.7 kcal, 52. (a) : H = (Ea)f 1– (Ea) b = 0, 53. (a) : C(graphite) + O2(g)  CO(g), 2, 1 1, n  1  , g, 2 2, As amount of gaseous substance is increasing in the, product side thus, S is positive for this reaction., And we know that G = H – TS, As S is positive, thus increase in temperature will make, the term (–TS) more negative and G will decrease., 54. (c) : Hfus = 1.435 kcal/mol, H fus 1.435 103, S , ,  5.26 cal/(mol K), T, fus, fus, 273, 55. (d) : We know that G = H – TS, 0 = H – TS, [ G = 0 as transition of, H2O(l), H2O(v) is at equilibrium], S , , H, , 3, , , , 30 10, ,  100 J mol1 K1, , T, 300, 56. (a) : Given reaction is :, 3, 1, 2 X2  2 Y2  XY3, We know, S°=, 3 – S°reactants,  1 S°products, (60)  (40) ,  50 , , , 2, 2, = 50 – (30 + 60) = – 40 J K–1 mol–1, At equilibrium G° = 0, H° = TS°, H 30  103 J mol1,  40 J K1 mol1  750 K, S, 57. (c) : According to Gibbs equation,, G = H – TS, when G = 0, H = TS, Given, H = 40.63 kJ mol–1 = 40.63 × 103 J mol–1, S = 108.8 J K–1 mol–1, H 40.63 103, T, ,  373.43 K, S, 108.8, 58. (b) : For the reaction to be spontaneous, G = –ve., Given : H = 170 kJ = 170 × 103 J, S = 170 J K–1, Applying, G = H – TS, the value of G = –ve only, when TS > H, which is possible only when T = 1110 K., T, , www.mediit.in

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Thermodynamics, , 53, ,  G = 170 × 103 – (1110 × 170) = –18700 J, Thus, reaction is spontaneous at T = 1110 K., , 68. (a) : Entropy change (S) is given by S , , 59. (d) : Gas phase reaction,, PCl5(g) PCl3(g) + Cl2(g), n H = E + ngRT, g = Change in number of moles of products and, reactants species., Since ng = +ve, hence H = +ve, also one mole of PCl5 is dissociated into two moles of, PCl3 and Cl2 in the same phase., Therefore, S = Sproducts – Sreactants, S = +ve., 60. (c) : The criteria for spontaneity of a process in, terms of G is as follows :, • If G is negative, the process is spontaneous., • If G is positive, the process does not occur in, the forward direction. It may occur in the backward, direction., • If G is zero, the system is in equilibrium., , q, 69. (d) : The change of entropy dS  rev, T, From the first law of thermodynamics,, dq = dU + PdV = C dT + PdV, , , , 61. (b) : Br2(l) + Cl2(g)  2BrCl(g), H = 30 kJ mol–1, S = 105 J K–1 mol–1, 30, H, i.e. 105  1000, S , T, T, 30 1000, T,  285.7 K, 105, 62. (a) : For spontaneous reaction, H = –ve, S = +ve., Spontaneity depends upon both critical minimum, energy and maximum randomness disorderness., 63. (c) : G = H – TS, G = –ve for spontaneous reaction., When S = +ve, H = +ve and TS > H  G = –ve, 64. (b) : G = H – TS, = 382.64  298  145.6 , 1000 , = – 382.64 + 43.38 = – 339.3 kJ mol–1, 65. (a) : For spontaneous process, Stotal > 0., , , Ssys + Ssurr > 0, q, 6000, 66. (d) : S  rev ,  21.978 J K1mol1, T, 273, 67. (a) : G° = – PV = Work done, 12 ,  12, V , , 103 L  1.71 103 L,  3.31 2.25 , G° = Work done = – (–1.71 × 10–3) × P × 101.3 J, 1895, 3, P, , 10.93, , 10, atm, 1.71  103 101.3, = 11.08 × 108 Pa  9.92 × 108 Pa ( 1 atm = 101325 Pa), , Unit of entropy = J K–1mol–1, , T, , V, , dT P,  CV,  dV, T, T T, dq, dT RdV,  C, , V, T, T, V, dT, dV,  dS  CV  R, T, V, T2, , , qrev, , dq, , , , For 1 mole of a gas,, , , , , P, T, , , , R, , , V , , V2, , [for one mole of ideal gas],  S  CV ln  R ln, T, T1, V1, Here T2 = T1 = 27°C, = 300, K  ln 2  0, T1, V2, 20,  S = R ln  2 ln = 2 ln 10 = 4.605, V1, 2,  S = 4.605 cal/mol K, Entropy change for 2 moles of gas, = 2 × 4.605 cal/K = 9.2 cal/K, 70. (d) : The sign and magnitude of Gibbs free energy is, a criterion of spontaneity for a process., When G > 0 or +ve, it means Gproducts > Greactants, as G = Gproducts – G reactants, the reaction will not take place spontaneously, i.e. the, reaction should be spontaneous in reverse direction., +4, , +2, , SnO2  SnO; G > 0, (more, favourable), , G < 0 or –ve, the reaction or change occurs spontaneously., +4, , +2, , PbO2  PbO; G < 0, (more, favourable), , 71. (a) : For a cell reaction to be spontaneous, G°, should be negative. As G° = –nFE°cell, so the value will, be –ve only when E°cell is +ve., 72. (c) : The entropy of a substance increases with, increase in temperature. However at absolute zero the, entropy of a perfectly crystalline substance is taken as, zero, which is also called as third law of thermodynamics., 73. (d) : (i) The given reaction is a combustion reaction,, therefore H is less than 0. Hence, H is negative., (ii) Since there is increase in the number of moles of, gaseous products, therefore S is positive., (iii) Since reaction is spontaneous, therefore G is, negative., , , , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 7, , Equilibrium, (b) 1.8 × 10–3, (c) 6.0 × 10–2, , 7.1 Equilibrium in Physical Processes, 1., , In liquid-gas equilibrium, the pressure of vapours, above the liquid is constant at, (a) constant temperature, (b) low temperature, (c) high temperature, (1995), , (d) none of these., , 7.3 Law of Chemical Equilibrium, Equilibrium Constant, 2., , 5., , 5, K, O2 2NO + 3H O will be, , 2, 2, (a) K K 3/K, (b) K K /K, , 6., , 2NH3 +, , 2 3, , 1, , 2 3, , (c) K23K 3/K 1, , 1, , (d) K1K33/K 2, (NEET 2017, 2007, 2003), , 3 . If the equilibrium constant for, N2(g) + O2(g), , 2NO(g) is K, the equilibrium, 1, constant for N2(g )  O2(g ) NO(g ) will be, 2, 2, 1, (b) K, (a) 2 K, 1, , (c) K2, 4., , (d) K1/2, , 7., , (2015), , Given that the equilibrium constant for the reaction,, 2 SO2(g )  O2(g ) 2 SO3(g ), has a value of 278 at a particular temperature. What, is the value of the equilibrium constant for the, following reaction at the same temperature?, 1,  O, SO, SO, 3(g ), , www.neetujee.com, , 2(g ), , 2, , 2(g ), , Given the reaction between 2 gases represented by, A2 and B2 to give the compound AB(g)., A2(g )  B2(g ) 2AB(g ), At equilibrium, the concentration of, A2 = 3.0 × 10–3 M, of B2 = 4.2 × 10–3 M, of, AB = 2.8 × 10–3 M, If the reaction takes place in a sealed vessel at 527°C,, then the value of Kc will be, (a) 2.0, (b) 1.9, (c) 0.62, (d) 4.5, (Mains 2012), , and, , The equilibrium constants of the following are, N2 + 3H2  2NH3; K1, N2 + O2  2NO; K2, 1, H2 + O2  H2O; K3, 2, The equilibrium constant (K) of the reaction :, , (b) 3.6 × 10–3, (d) 1.3 × 10–5, (Mains 2012), , 8., , For the reaction, N 2(g) + O 2(g)  2NO (g) , the, equilibrium constant is K1. The equilibrium constant, is K2 for the reaction,, 2NO(g) + O2(g)  2NO2(g), What is K for the reaction,, 1, NO, N +O ?, 2(g) , 2(g), 2(g), 2, 1, 1, (b), (a), 4K1K2, 2K1K2, 1/2, , 1,  1, (2011), (c) , , (d) K K, KK,  1 2, 12, The dissociation constants for acetic acid and HCN, at 25°C are 1.5 × 10–5 and 4.5 × 10–10 respectively., The equilibrium constant for the equilibrium,, CN– + CH3COOH  HCN + CH3COO–would be, (a) 3.0 × 10–5, (b) 3.0 × 10–4, 4, (c) 3.0 × 10, (d) 3.0 × 105, (2009), The value of, 1 equilibrium, 1 constant of the reaction,, HI (g )  H2 (g )  I2(g ), 2, 2, is 8.0. The equilibrium constant of the reaction, H2( g )  I2(g )  2HI (g ) will be, , www.mediit.in

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Equilibrium, , 9., , 55, , (a) 16, (b) 1/8, (c) 1/16, (d) 1/64, (2008), Equilibrium constants K1 and K2 for the following, equilibria1:, K, 1 NO, NO  O, and, (g ), , 2NO, , 2(g ), , 2( g ), , 2, K2, , 2NO, , 2(g ), , O, (g ), , 2(g ), , (a) (2Kp/P) 1/2, , (b) (Kp/P), , (c) (2Kp/P), , (d) (2Kp/P)1/3, 1, , (b) K = K 2, (2005), (d) 2 = K 1, K2 1/2, 10. If K1 and K2 are the respective equilibrium constants, for the two reactions,, XeF6(g) + H2O(g)  XeOF4(g) + 2HF(g), XeO4(g) + XeF6(g)  XeOF4(g) + XeO3F2(g),, the equilibrium constant of the reaction,, XeO4(g) + 2HF(g)  XeO3F2(g) + H2O(g),, will be, (a) K1/K2, (b) K1·K2, 2, , (c) K1/(K2), (d) K2/K1, (1998), 11. The equilibrium constant for the reaction, N2 + 3H2  2NH3 is K, then the equilibrium, constant for the equilibrium 2NH3  N2 + 3H2is, 1, 1, (c) 1, (a) K, (b), (d) 2, K, K, K, (1996), 12. K1 and K2 are equilibrium constants for reactions (i), and (ii) respectively., N2(g) + O2(g), 2NO(g) ........................................................(i), 1, 1, NO, N + O, (g), 2(g), 2(g), 2, ...(ii), 2, 2, 1, (b) K1 = K 22, (a) K1   , K2 , (c) K  1, (d) K1 = (K2) 0, (1989), 1 K, 2, , 7.4 Homogeneous Equilibrium, 13. The reaction, 2A(g) + B(g)  3C(g) + D(g), is begun with the concentrations of A and B both, at an initial value of 1.00 M. When equilibrium is, reached, the concentration of D is measured and, found to be 0.25 M. The value for the equilibrium, constant for this reaction is given by the expression, (a) [(0.75)33(0.25)]  [(1.00)22(1.00)], (b) [(0.75) (0.25)]  [(0.50) (0.75)], (c) [(0.75)3(0.25)]  [(0.50)2(0.25)], (d) [(0.75)3(0.25)]  [(0.75)2(0.25)] (Mains 2010), , www.neetujee.com, , The degree of dissociation is x and is small, compared to 1. The expression relating the degree, of dissociation (x) with equilibrium constant Kp and, total pressure P is, (2008), , 15. The values of Kp and Kp for the reactions,, , , are related as, (a) K2 = 1/K 2, (c) K = 1/ 1, K1, 2, , 14. The dissociation equilibrium of a gas AB2, represented as :, , 2AB2(g)  2AB(g) + B2(g), , can be, , 2, , X  Y+ Z, A  2B, , ...(i), ...(ii), , are in the ratio 9 : 1. If degree of dissociation of X, and A be equal, then total pressure at equilibrium (i), and (ii) are in the ratio, (a) 36 : 1, (b) 1 : 1, (c) 3 : 1, (d) 1 : 9, (2008), , 7.5 Heterogeneous Equilibrium, 16. A 20 litre container at 400 K contains CO2(g) at, pressure 0.4 atm and an excess of SrO (neglect the, volume of solid SrO). The volume of the container, is now decreased by moving the movable piston, fitted in the container. The maximum volume of, the container, when pressure of CO2 attains its, maximum value, will be, (Given that : SrCO3(s) SrO(s) + CO2(g),, Kp = 1.6 atm), (a) 10 litre, (b) 4 litre, (c) 2 litre, (d) 5 litre (NEET 2017), 17. In which of the following equilibrium Kc and Kp are, not equal?, (a) 2NO(g)  N2(g) + O2(g), (b) SO2(g) + NO2(g)  SO3(g) + NO(g), (c) H2(g) + I2(g)  2HI(g), (d) 2C (s) + O2(g)  2CO2(g), (2010), 18. If the concentration of OH– ions in the reaction, 3+, –, Fe(OH)3(s)  Fe(aq, ) + 3OH, (aq ), is decreased by 1/4 times, then equilibrium, concentration of Fe3+ will increase by, (a) 64 times, (b) 4 times, (c) 8 times, (d) 16 times., (2008), 19. Equilibrium constant Kp for following reaction, MgCO3(s) MgO(s) + CO2(g), (a) Kp = pCO2, (b) Kp = pCO2 ×, (c) K , p, , pCO2  pMgO, pMgCO3, , pCO2  pMgO, pMgCO3, , (d) K , p, , pMgCO3, , pCO2  pMgO, (2000), www.mediit.in

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56, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 7.6 Applications of Equilibrium Constant, 20. If the value of equilibrium constant for a particular, reaction is 1.6 × 1012, then at equilibrium the system, will contain, (a) mostly products, (b) similar amounts of reactants and products, (c) all reactants, (d) mostly reactants., (2015, Cancelled), 21. In Haber process, 30 litres of dihydrogen and 30, litres of dinitrogen were taken for reaction which, yielded only 50% of the expected product. What, will be the composition of gaseous mixture under, the aforesaid condition in the end?, (a) 20 litres ammonia, 20 litres nitrogen, 20 litres, hydrogen, (b) 10 litres ammonia, 25 litres nitrogen, 15 litres, hydrogen, (c) 20 litres ammonia, 10 litres nitrogen, 30 litres, hydrogen, (d) 20 litres ammonia, 25 litres nitrogen, 15 litres, hydrogen, (2003), 22. The reaction quotient (Q) for the reaction, N2(g) + 3H2(g), 2NH3(g) is given by, 2, , [NH3 ], , . The reaction will proceed from, [N2 ] [H2 ]3, right to left if, (b) Q < Kc, (a) Q = Kc, (d) Q = 0, (c) Q > K, (2003), where K c, c is the equilibrium constant., Q=, , 7.7 Relationship Between K, Q, and G, 23. Hydrolysis of sucrose is given by the following, reaction : Sucrose + H2O, Glucose + Fructose, If the equilibrium constant (KC) is 2 × 1013 at 300 K,, the value of rG° at the same temperature will be, (a) –8.314 J mol–1K–1 × 300 K × ln(2 × 1013), (b) 8.314 J mol–1K–1 × 300 K × ln(2 × 1013), (c) 8.314 J mol–1K–1 × 300 K × ln(3 × 1013), (d) –8.314 J mol–1K–1 × 300 K × ln(4 × 1013), (NEET 2020), , List I, List II, (Equations), (Type of processes), A. Kp > Q, (i) Non- spontaneous, B. G° < RT ln Q (ii) Equilibrium, C. Kp = Q, (iii) Spontaneous, and endothermic, H, D. T , (iv) Spontaneous, S, (a) A - (i), B - (ii), C - (iii), D - (iv), (b) A - (iii), B - (iv), C - (ii), D - (i), (c) A - (iv), B - (i), C - (ii), D - (iii), (d) A - (ii), B - (i), C - (iv), D - (iii) (Mains 2010), , 7.8 Factors Affecting Equilibrium, 26. Which one of the following conditions will favour, maximum formation of the product in the reaction, A2(g) + B2(g)  X2(g) , rH = –X kJ ?, (a) Low temperature and high pressure, (b) Low temperature and low pressure, (c) High temperature and high pressure, (d) High temperature and low pressure, (NEET 2018), 27. For the reversible reaction,, N2(g) + 3H2(g) 2NH3(g) + heat, The equilibrium shifts in forward direction, (a) by increasing the concentration of NH3(g), (b) by decreasing the pressure, (c) by decreasing the concentrations of N 2(g) and, H2(g), (d) by increasing pressure and decreasing, temperature., (2014), 28. For a given exothermic reaction, Kp and Kp are, the equilibrium constants at temperatures T1 and, T2, respectively. Assuming that heat of reaction is, constant in temperature range between T1 and T2, it, is readily observed that, (a) Kp > Kp, , (b) Kp < Kp, , (c) Kp = Kp, , (d) Kp , , , Kp, , (2014), , 24. Which of the following statements is correct for a, reversible process in a state of equilibrium?, (a) G° = –2.30 RT log K (b) G° = 2.30 RT log K, (c) G = –2.30 RT log K (d) G = 2.30 RT log K, (2015, Cancelled), , 29. KMnO4 can be prepared from K2MnO4 as per the, reaction,, –, 3MnO42– + 2H2 O 2MnO – + MnO, 4, 2 + 4OH, The reaction can go to completion by removing, OH– ions by adding, (a) CO2, (b) SO2, (c) HCl, (d) KOH, (NEET 2013), , 25. Match List I (Equations) with List II (Type of, processes) and select the correct option., , 30. The value of H for the reaction, X2(g) + 4Y2(g)  2XY4(g) is less than zero., , www.neetujee.com, , www.mediit.in

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Equilibrium, , 57, , Formation of XY4(g) will be favoured at, (a) high temperature and high pressure, (b) low pressure and low temperature, (c) high temperature and low pressure, (d) high pressure and low temperature., , (a), (b), (c), (d), (2011), , 31. For the reaction :, CH4(g) + 2O2(g) CO2(g) + 2H2O(l),, Hr = –170.8 kJ mol–1. Which of the following, statements is not true?, (a) The reaction is exothermic., (b) At equilibrium, the concentrations of CO2(g), and H2O(l) are not equal., (c) The equilibrium constant for the reaction is, [CO2 ], given by K, ., p, [CH ][O ], 4, , 2, , (d) Addition of CH4(g) or O2(g) at equilibrium will, cause a shift to the right., (2006), 32. Reaction BaO2(s), BaO(s) + O2(g); H = +ve. In, , Hydrochloric acid, HCl, Ammonia, NH 3, Fructose, C6H12O6, Acetic acid, C2H4O2, , 37. Aqueous solution of acetic acid contains, (a) CH3 COO– and H +, (b) CH3 COO– , H 3O+ and CH 3COOH, +, (c) CH3 COO– , H 3O+ and H, (d) CH3COOH, CH3COO– and H+, , 33. For any reversible reaction, if we increase, concentration of the reactants, then effect on, equilibrium constant, (a) depends on amount of concentration, (b) unchange, (c) decrease, (d) increase., (2000), 34. According to Le Chatelier’s principle, adding heat to, a solid and liquid in equilibrium will cause the, (a) temperature to increase, (b) temperature to decrease, (c) amount of liquid to decrease, (d) amount of solid to decrease., (1993), 35. Which one of the following information can be, obtained on the basis of Le Chatelier principle?, (a) Dissociation constant of a weak acid, (b) Entropy change in a reaction, (c) Equilibrium constant of a chemical reaction, (d) Shift in equilibrium position on changing value, of a constraint, (1992), , 7.9 Ionic Equilibrium in Solution, 36. Aqueous solution of which of the following, compounds is the best conductor of electric current?, , www.neetujee.com, , (1991), , 7.10 Acids, Bases and Salts, 38. Conjugate base for Bronsted acids H O and HF are, 2, (a) H O+ and H F+, respectively, 3, , 2, , –, , +, , (b) OH and H2F , respectively, (c) H3O+ and F–, respectively, (d) OH– and F–, respectively., , (NEET 2019), , 39. Which of the following cannot act both as Bronsted, acid and as, Bronsted base ?, (a) HCO–, (b) NH, 3, , equilibrium condition, pressure of O2 depends on, (a) increase mass of BaO2, (b) increase mass of BaO, (c) increase temperature on equilibrium, (d) increase mass of BaO2 and BaO both., (2002), , (2015), , (c) HCl, , 3, , –, , (d) HSO 4, (Odisha NEET 2019), , 40. Which of the following fluoro-compounds is most, likely to behave as a Lewis base?, (a) BF3, (b) PF3, (c) CF4, (d) SiF4, (NEET-II 2016), 41. Which of these is least likely to act as a Lewis base?, (a) BF3, (b) PF3, (c) CO, (d) F–, (NEET 2013), 42. Which is the strongest acid in the following?, (a) HClO4, (b) H2SO3, (c) H2SO4, (d) HClO3 (NEET 2013), 43. Which one of the following molecular hydrides acts, as a Lewis acid?, (a) NH3, (b) H2O, (c) B2H6, (d) CH4, (2010), 44. Which of the following molecules acts as a Lewis, acid?, (a) (CH3)2O, (b) (CH3)3P, (c) (CH3)3N, (d) (CH3)3B, (2009), 45. Which one of the following statements is not true?, (a) Among halide ions, iodide is the most powerful, reducing agent., (b) Fluorine is the only halogen that does not show, a variable oxidation state., (c) HOCl is a stronger acid than HOBr., (d) HF is a stronger acid than HCl., (2003), 46. Which one of the following compounds is not a, protonic acid?, , www.mediit.in

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58, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (a) B(OH) 3, (b) PO(OH) 3, (c) SO(OH)2, (d) SO2(OH)2, (2003), – –, 47. In HS , I , R – NH2, NH3 order of proton accepting, tendency will be, (a) I– > NH3 > R – NH2 > HS–, (b) NH3 > R – NH2 > HS– > I–, (c) R – NH2 > NH3 > HS– > I–, (d) HS– > R – NH2 > NH3 > I–, (2001), 48. Conjugate acid of NH– is, , pH of which one of them will be equal to 1?, (a) B, (b) A, (c) D, (d) C, (NEET 2018), , (a) NH 4OH, (c) NH 2–, , 2, , (b) NH + 4, (d) NH3, , 49. Which compound is electron deficient?, (a) BeCl2, (b) BCl3, (c) CCl4, (d) PCl5, 50. The strongest, conjugate base is –, (a) SO 2–, (b) Cl, 4, , –, , (c) NO3, , (d) CH3COO–, , (2000), , (2000), , (1999), , 51. Which of the following is not a Lewis acid?, (a) SiF4, (b) C2H4, (1996), (c) BF3, (d) FeCl3, 52. Repeated use of which one of the following fertilizers, would increase the acidity of the soil?, (a) Ammonium sulphate, (b) Superphosphate of lime, (c) Urea, (d) Potassium nitrate, (1998), , 7.11 Ionization of Acids and Bases, , 56. The percentage of pyri+dine (C5H5N) that forms, pyridinium ion (C5H5NH) in a 0.10 M aqueous, pyridine solution (Kb for C5H5N = 1.7 × 10–9) is, (a) 0.0060%, (b) 0.013%, (c) 0.77%, (d) 1.6% (NEET-II 2016), 57. What is the pH of the resulting solution when equal, volumes of 0.1 M NaOH and 0.01 M HCl are mixed?, (a) 2.0, (b) 7.0, (c) 1.04, (d) 12.65, (2015), 58. Which of the following salts will give highest pH in, water?, (a) KCl, (b) NaCl, (c) Na2CO3, (d) CuSO4, (2014), 59. Accumulation of lactic acid (HC3H5O3), a, monobasic acid in tissues leads to pain and a feeling, of fatigue. In a 0.10 M aqueous solution, lactic acid is, 3.7% dissociated. The value of dissociation constant,, Ka, for this acid will be, (a) 1.4 × 10–5, (b) 1.4 × 10–4, –4, (c) 3.7 × 10, (d) 2.8 × 10–4, (Karnataka NEET 2013), 60. At 100°C the Kw of water is 55 times its value at, 25°C. What will be the pH of neutral solution?, (log 55 = 1.74), (a) 7.00 (b) 7.87 (c) 5.13 (d) 6.13, (Karnataka NEET 2013), , 53. Find out the solubility of Ni(OH)2 in 0.1 M NaOH., Given that the ionic product of Ni(OH)2 is 2 × 10–15., (a) 2 × 10–13 M, (b) 2 × 10–8 M, (c) 1 × 10–13 M, (d) 1 × 108 M, (NEET 2020), , 61. Equimolar solutions of the following substances, were prepared separately. Which one of these will, record the highest pH value?, (a) BaCl2, (b) AlCl3, (c) LiCl, (d) BeCl2, (2012), , 54. The pH of 0.01 M NaOH(aq) solution will be, (a) 7.01, (b) 2, (c) 12, (d) 9, (Odisha NEET 2019), , 62. What is [H+] in mol/L of a solution that is 0.20 M, in CH3COONa and 0.10 M in CH3COOH?, (Ka for CH3COOH = 1.8 × 10–5), (a) 3.5 × 10–4, (b) 1.1 × 10–5, –5, (c) 1.8 × 10, (d) 9.0 × 10–6, (2010), , 55. Following solutions were prepared by mixing, different volumes of NaOH and HCl of different, concentrations :, A. 60 mL M HCl  40 mL M NaOH, 10, 10, B. 55 mL M HCl  45 mL M NaOH, 10, 10, M, M, C. 75 mL HCl  25 mL NaOH, 5, 5, M, D. 100 mL HCl  100 mL M NaOH, 10, 10, , www.neetujee.com, , 63. The ionization constant of ammonium hydroxide, is 1.77 × 10–5 at 298 K. Hydrolysis constant of, ammonium chloride is, (a) 6.50 × 10–12, (b) 5.65 × 10–13, –12, (c) 5.65 × 10, (d) 5.65 × 10–10, (2009), 64. What is the [OH–] in the final solution prepared by, mixing 20.0 mL of 0.050 M HCl with 30.0 mL of, 0.10 M Ba(OH)2?, (a) 0.40 M, (b) 0.0050 M, (c) 0.12 M, (d) 0.10 M, (2009), , www.mediit.in

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Equilibrium, , 59, , 65. Equal volumes of three acid solutions of pH 3, 4, and 5 are mixed in a vessel. What will be the H+ ion, concentration in the mixture?, (a) 3.7 × 10–3 M, (b) 1.11 × 10–3 M, –4, (c) 1.11 × 10 M, (d) 3.7 × 10–4 M (2008), –5, , 66. A weak acid, HA, has a Ka of 1.00 × 10 . If 0.100, mol of this acid is dissolved in one litre of water,, the percentage of acid dissociated at equilibrium is, closest to, (a) 1.00%, (b) 99.9%, (c) 0.100%, (d) 99.0%, (2007), 67. Calculate the pOH of a solution at 25°C that contains, 1 × 10–10 M of hydronium ions, i.e. H3O+., (a) 4.000, (b) 9.000, (c) 1.000, (d) 7.000, (2007), 68. Thehydrogen ion concentration ofa 10–8 M HCl aqueous, solution at 298 K (Kw = 10–14) is, (a) 1.0 × 10–8 M, (b) 1.0 × 10–6 M, (c) 1.0525 × 10–7 M, (d) 9.525 × 10–8 M (2006), 69. At 25°C, the dissociation constant of a base, BOH,, is 1.0 × 10–12. The concentration of hydroxyl ions in, 0.01 M aqueous solution of the base would be, (a) 1.0 × 10–5 mol L–1, (b) 1.0 × 10–6 mol L–1, –6, –1, (c) 2.0 × 10 mol L, (d) 1.0 × 10–7 mol L–1 (2005), 70. Which has highest pH?, (a) CH3COOK, (b) Na2CO3, (c) NH4Cl, (d) NaNO3, (2002), 71. Ionisation constant of CH3COOH is 1.7 × 10–5, and concentration of H+ ions is 3.4 × 10–4. Then find, out initial concentration of CH3COOH molecules., (a) 3.4 × 10–4, (b) 3.4 × 10–3, (2001), (c) 6.8 × 10–4, (d) 6.8 × 10–3, 72. Correct relation between dissociation constants of a, dibasic acid is, (a) Ka1  Ka 2, (b) Ka1  Ka 2, (c) Ka1  Ka 2, , 1, (d) Ka1  K, a2, , (2000), , 73. Which statement is wrong about pH and H+?, (a) pH of neutral water is not zero., (b) Adding 1 N solution of CH3COOH and 1 N, solution of NaOH, pH will be seven., (c) [H+] of dilute and hot H2SO 4 is more than, concentrated and cold H2SO4., (d) Mixing solution of CH3COOH and HCl, pH, will be less than 7., (2000), 74. The concentration of [H+] and concentration of, [OH–] of a 0.1 aqueous solution of 2% ionised weak, acid is [ionic product of water = 1 × 10–14], , www.neetujee.com, , (a) 2 × 10–3 M and 5 × 10–12 M, (b) 1 × 10–3 M and 3 × 10–11 M, (c) 0.02 × 10–3 M and 5 × 10–11 M, (d) 3 × 10–2 M and 4 × 10–13 M, , (1999), , –, , 75. The hydride ion H is stronger base than its, hydroxide ion OH–. Which of the following reaction, will occur if sodium hydride (NaH) is dissolved in, water?, (a) H– + H2O  no reaction, (b) H––(aq) + H2O  H2 O, –, (c) H (aq) + H2O(l)  OH + H 2, (1997), , (d) None of these., –14, , 76. The ionic product of water at 25°C is 10 . Its ionic, product at 90°C will be,, (a) 1 × 10–14, (b) 1 × 10–16, (c) 1 × 10–20, (d) 1 × 10–12, (1996), 77. If  is dissociation constant, then the total number, of moles for the reaction, 2HI  H2 + I2 will be, (a) 1, (b) 1 – , (c) 2, (d) 2 – , (1996), 78. The pH value of N/10 NaOH solution is, (a) 12 (b) 13 (c) 10 (d) 11 (1996), 79. The pH value of a 10 M solution of HCl is, (a) equal to 1, (b) equal to 2, (c) less than 0, (d) equal to 0 (1995), 80. At 80°C, distilled water has [H3O+] concentration, equal to 1 × 10–6 mole/litre. The value of Kw at this, temperature will be, (a) 1 × 10–12, (b) 1 × 10–15, (c) 1 × 10–6, (d) 1 × 10–9, (1994), 81. 0.1 M solution of which one of these substances will, act basic?, (a) Sodium borate (b) Ammonium chloride, (c) Calcium nitrate (d) Sodium sulphate, (1992), 82. The compound whose water solution has the highest, pH is, (a) NaCl, (b) NaHCO3, (c) Na2CO3, (d) NH4Cl, (1988), , 7.12 Buffer Solutions, 83. Which will make basic buffer?, (a) 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH, (b) 50 mL of 0.1 M NaOH + 25 mL of 0.1 M, CH3COOH, (c) 100 mL of 0.1 M CH3COOH + 100 mL of 0.1 M, NaOH, (d) 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH4OH, (NEET 2019), , www.mediit.in

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60, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 84. Which one of the following pairs of solutions is not, an acidic buffer?, (a) CH 3 COOH and CH 3COONa, (b) H2CO 3 and Na 2CO3, (c) H3PO 4 and Na 3PO4, (d) HClO4 and NaClO4, (2015), 85. The dissociation constant of a weak acid is 1 × 10–4., In order to prepare a buffer solution with a pH = 5,, the [Salt]/[Acid] ratio should be, (a) 4 : 5 (b) 10 : 1 (c) 5 : 4 (d) 1 : 10, (Karnataka NEET 2013), 86. Buffer solutions have constant acidity and alkalinity, because, (a) these give unionised acid or base on reaction, with added acid or alkali, (b) acids and alkalies in these solutions are shielded, from attack by other ions, (c) they have large excess of H+ or OH– ions, (d) they have fixed value of pH., (2012), 87. A buffer solution is prepared in which the, concentration of NH 3 is 0.30 M and the, concentration of NH4+ is 0.20 M. If the equilibrium, constant, K b for NH3 equals 1.8 × 10 –5, what is the, pH of this solution? (log 2.7 = 0.43), (a) 9.08, (b) 9.43, (c) 11.72, (d) 8.73, (2011), 88. In a buffer solution containing equal concentration, of B– and HB, the Kb for B– is 10–10. The pH of buffer, solution is, (a) 10, (b) 7, (c) 6, (d) 4 (2010), 89. Which of the following pairs constitutes a buffer?, (a) HCl and KCl, (b) HNO2 and NaNO2, (c) NaOH and NaCl, (d) HNO 3 and NH4NO 3, (2006), 90. The rapid change of pH near the stoichiometric, point of an acid-base titration is the basis of indicator, detection. pH of the solution is related to ratio of the, concentrations of the conjugate acid (HIn) and base, (In–) forms of the indicator by the expression, [In ],  pKIn  pH, [HIn], [HIn], (b) log,  pK In  pH, , [In ], [HIn], (c) log,  pH  pKIn , [In ], [In ], (d) log,  pH  pK, In, [HIn], (a) log, , www.neetujee.com, , (2004), , 91. Solution of 0.1 N NH4OH and 0.1 N NH4Cl has pH, 9.25. Then find out pKb of NH4OH., (a) 9.25, (b) 4.75, (c) 3.75, (d) 8.25, (2002), 92. A physician wishes to prepare a buffer solution at, pH = 3.85 that efficiently resists changes in pH yet, contains only small concentration of the buffering, agents. Which of the following weak acids together, with its sodium salt would be best to use?, (a) 2, 5-Dihydroxybenzoic acid (pKa = 2.97), (b) Acetoacetic acid (pKa = 3.58), (c) m-Chlorobenzoic acid (pKa = 3.98), (d) p-Chlorocinnamic acid (pKa = 4.41), (1997), 93. The pH value of blood does not appreciably change, by a small addition of an acid or a base, because the, blood, (a) can be easily coagulated, (b) contains iron as a part of the molecule, (c) is a body fluid, (d) contains serum protein which acts as buffer., (1995), , 7.13 Solubility Equilibrium, Soluble Salts, , of, , Sparingly, , 94. pH of a saturated solution of Ca(OH)2 is 9. The, solubility product (Ksp) of Ca(OH)2 is, (a) 0.5 × 10–10, (b) 0.5 × 10–15, –10, (c) 0.25 × 10, (d) 0.125 ×10–15, (NEET 2019), 95. The molar solubility of CaF2 (Ksp = 5.3 × 10–11) in 0.1, M solution of NaF will be, (a) 5.3 × 10–11 mol L–1 (b) 5.3 × 10–8 mol L–1, (c) 5.3 × 10–9 mol L–1, (d) 5.3 × 10–10 mol L–1, (Odisha NEET 2019), 96. The solubility of BaSO4 in water is 2.42 × 10–3 g L–1, at 298 K. The value of its solubility product (Ksp) will, be, (Given molar mass of BaSO4 = 233 g mol–1), (a) 1.08 × 10–10 mol2 L–2, (b) 1.08 × 10–12 mol2 L–2, (c) 1.08 × 10–14 mol2 L–2, (d) 1.08 × 10–8 mol2 L–2, (NEET 2018), 97. Concentration of the Ag+ ions in a saturated solution, of Ag2C2O4 is 2.2 × 10–4 mol L–1. Solubility product, of Ag2C2O4 is–12, (a) 2.66 × 10, (b) 4.5 × 10–11, –12, (c) 5.3 × 10, (d) 2.42 × 10–8, (NEET 2017), , www.mediit.in

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Equilibrium, , 61, , 98. The solubility of AgCl(s) with solubility product, 1.6 × 10–10 in 0.1 M NaCl solution would be, (a) 1.26 × 10–5 M, (b) 1.6 × 10–9 M, (c) 1.6 × 10–11 M, (d) zero. (NEET-II 2016), 99. MY and NY , two nearly insoluble salts, have the, 3, –13, same K, sp values of 6.2 × 10, at room temperature., Which statement would be true in regard to MY and, NY3?, (a) The salts MY and NY3 are more soluble in 0.5 M, KY than in pure water., (b) The addition of the salt of KY to solution of MY, and NY3 will have no effect on their solubilities., (c) The molar solubilities of MY and NY3 in water, are identical., (d) The molar solubility of MY in water is less than, (NEET-I 2016), that of NY3., 100. The K of Ag CrO , AgCl, AgBr and AgI are, sp, , 2, , 4, , respectively, 1.1 × 10–12, 1.8 × 10–10, 5.0 × 10–13,, 8.3 × 10–17. Which one of the following salts will, precipitate last if AgNO3 solution is added to the, solution containing equal moles of NaCl, NaBr, NaI, and Na2CrO4?, (a) AgBr, (b) Ag2CrO4, (c) AgI, (d) AgCl, (2015, Cancelled), 101. Using the Gibbs’ energy change, G° = +63.3 kJ, for, the following reaction,, Ag2CO3(s), 2Ag+(aq) + CO 32–( aq), the Ksp of Ag2CO3(s) in water at 25 °C is, (R = 8.314 J K–1 mol–1), (a) 3.2 × 10–26, (b) 8.0 × 10–12, –3, (c) 2.9 × 10, (d) 7.9 × 10–2, (2014), 102. The values of Ksp of CaCO3 and CaC 2O4 are, –9, , –9, , 4.7 × 10 and 1.3 × 10 respectively at 25°C. If the, mixture of these two is washed with water, what is, the concentration of Ca2+ ions in water?, (a) 5.831 × 10–5 M, (b) 6.856 × 10–5 M, –5, (c) 3.606 × 10 M, (d) 7.746 × 10–5 M, (Karnataka NEET 2013), 103. Identify the correct order of solubility in aqueous, medium., (a) Na2S > CuS > ZnS (b) Na2S > ZnS > CuS, (c) CuS > ZnS > Na2S (d) ZnS > Na2S > CuS, (NEET 2013), 104. pH of a saturated solution of Ba(OH)2 is 12. The, value of solubility product (Ksp) of Ba(OH)2 is, (a) 3.3 × 10–7, (b) 5.0 × 10–7, (c) 4.0 × 10–6, (d) 5.0 × 10–6 (2012, 2010), , www.neetujee.com, , 105. In qualitative analysis, the, metals, of, group I can be separated from other ions by, precipitating them as chloride salts. A solution, initially contains Ag+ and Pb2+ at a concentration, of 0.10 M. Aqueous HCl is added to this solution, until the Cl– concentration is 0.10 M. What will the, concentrations of Ag+ and Pb2+ be at equilibrium?, (Ksp for AgCl = 1.8 × 10–10, Ksp for PbCl2, = 1.7 × 10–5), +, –7, 2+, (a) [Ag ] = 1.8 × 10 M, [Pb ] = 1.7 × 10–6 M, (b) [Ag+] = 1.8 × 10–11 M, [Pb2+] = 8.5 × 10–5 M, (c) [Ag+] = 1.8 × 10–9 M, [Pb2+] = 1.7 × 10–3 M, (d) [Ag+] = 1.8 × 10–11 M, [Pb2+] = 1.7 × 10–4 M, (Mains 2011), 106. H2S gas when passed through a solution of cations, containing HCl precipitates the cations of second, group of qualitative analysis but not those belonging, to the fourth group. It is because, (a) presence of HCl decreases the sulphide ion, concentration, (b) solubility product of group II sulphides is more, than that of group IV sulphides, (c) presence of HCl increases the sulphide ion, concentration, (d) sulphides of group IV cations are unstable in, HCl., (2005), 107. The solubility product of a sparingly soluble salt AX2, is 3.2 × 10–11. Its solubility (in moles/L) is, (a) 5.6 × 10–6, (b) 3.1 × 10–4, –4, (c) 2 × 10, (d) 4 × 10–4, (2004), 108. The solubility product of AgI at 25°C is, 1.0 × 10–16 mol2 L–2. The solubility of AgI in 10–4 N, solution of KI at 25°C is approximately (in mol L–1), (a) 1.0 × 10–16, (b) 1.0 × 10–12, (c) 1.0 × 10–10, , (d) 1.0 × 10–8, , (2003), , 109. Solubility of MX2 type electrolytes is 0.5 × 10–4, mol/lit., then find out Ksp of electrolytes., (a) 5 × 10–12, (b) 25 × 10–10, –13, (c) 1 × 10, (d) 5 × 10–13, (2002), 110. Solubility of M2S salt is 3.5 × 10–6 then find out, solubility product., (a) 1.7 × 10–6, (b) 1.7 × 10–16, (c) 1.7 × 10–18, (d) 1.7 × 10–12, (2001), 111. The solubility of a saturated solution of calcium, fluoride is 2 × 10–4 moles per litre. Its solubility, product is, (a) 22 × 10–11, (b) 14 × 10–4, –2, (c) 2 × 10, (d) 32 × 10–12, (1999), , www.mediit.in

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62, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 112. The solubility product of CuS, Ag2S and HgS are, 10–31, 10–44 and 10–54 respectively. The solubilities of, these sulphides are in the order, (a) HgS > Ag2S > CuS (b) CuS > Ag2S > HgS, , (a) 0.01 M CaCl2, (c) 0.001 M AgNO3, , (c) Ag2S > CuS > HgS (d) Ag2S > HgS > CuS, (1997), , (b) pure water, (d) 0.01 M NaCl, , (1995), , 114. Which one of the following is most soluble?, (a) Bi2S3 (Ksp = 1 × 10–70), (b) Ag2S (Ksp = 6 × 10–51), (c) CuS (Ksp = 8 × 10–37), (d) MnS (Ksp = 7 × 10–16), , 113. The solubility of AgCl will be minimum in, , (1994), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., 61., 71., 81., 91., 101., 111., , (a), (c), (b), (c), (a), (b), (a), (d), (a), (b), (b), (d), , 2., 12., 22., 32., 42., 52., 62., 72., 82., 92., 102., 112., , (a), (a), (c), (c), (a), (a), (d), (b), (c), (b), (d), (b), , 3., 13., 23., 33., 43., 53., 63., 73., 83., 93., 103., 113., , (d), (b), (a), (b), (c), (a), (d), (b), (d), (d), (b), (a), , 4., 14., 24., 34., 44., 54., 64., 74., 84., 94., 104., 114., , (c), (d), (a), (d), (d), (c), (d), (a), (d), (b), (b), (d), , 5., 15., 25., 35., 45., 55., 65., 75., 85., 95., 105., , (c), (a), (c), (d), (d), (d), (d), (c), (b), (c), (c), , 6., 16., 26., 36., 46., 56., 66., 76., 86., 96., 106., , (c), (d), (a), (a), (a), (b), (a), (d), (a), (a), (a), , 7., 17., 27., 37., 47., 57., 67., 77., 87., 97., 107., , (c), (d), (d), (b), (c), (d), (a), (c), (b), (c), (c), , 8., 18., 28., 38., 48., 58., 68., 78., 88., 98., 108., , (d), (a), (a), (d), (d), (c), (c), (b), (d), (b), (b), , 9., 19., 29., 39., 49., 59., 69., 79., 89., 99., 109., , (a), (a), (a), (c), (b), (b), (d), (c), (b), (d), (d), , 10., 20., 30., 40., 50., 60., 70., 80., 90., 100., 110., , (d), (a), (d), (b), (d), (d), (b), (a), (d), (b), (b), , Hints & Explanations, 1. (a) : Vapour pressure is directly related to, temperature. Greater is the temperature, greater will be, the vapour pressure. So to keep it constant, temperature, should be constant., 2. (a) : From the given equations,, 1, 2NH3  N2 + 3H2;, ...(i), K1, ...(ii), N2 + O2  2NO; K2, 3, 3, 3H2 + O2  3H2O; K3, ...(iii), 2, By adding equations (i), (ii) and (iii), we get 3, 5, K, K K, 2NH3  O2, 2NO+ 3H2O, K  2 3, 2, K1, , 3 2, , 1, 3. (d) : If the reaction is multiplied by , then new, 2, equilibrium constant, K = K1/2., 4. (c) : 2SO2(g) + O2(g)  2SO3(g), K = 278 .......... (i), By reversing the equation (i), we get, 2SO3(g)  2SO2(g) + O2(g), Equilibrium constant for this reaction is,, , www.neetujee.com, , 1 1, K , K 278, By dividing the equation (ii) by 2, we get desired equation,, 1, SO3(g ) SO2(g )  O2(g ), ...(iii), 2, Equilibrium constant for this reaction,, 1  1 = 0.0599 ≈ 0.06 or 6 × 10–2, K   K , K, 278, 5. (c) : A2(g) + B2(g)  2AB(g), Kc  [AB]2, [A2 ][B2 ], , ...(ii), , 6., , (2.8 10 ), 2.8  2.8, , ,  0.62, 3, 3, (3.0 10 )(4.2 10 ) 3.0  4.2, (c) : N2 + O2  2NO ; K1, , 2NO + O 2  2NO 2 ; K2, 1, NO2  N2 + O2; K, 2, 2, 2, [NO], K1 , ; K2  [NO2 ], [N2 ][O2 ], [NO] 2[O2 ], , www.mediit.in

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64, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , KP1, , 9 P1, P1 36, , , , P2  1  36 :1, 2, 1 P, 4 2, K p2, P2, 16. (d) : SrCO3(s)  SrO(s) + CO2(g); Kp = 1.6 am, t, pCO  pSrO, 2, Kp ,  1.6 = pCO ( pSrO = pSrCO = 1), 2, 3, pSrCO, Now, , Kp 1, ,  P1 , 4P K, , 3, ,  Maximum pressure of CO 2 = 1.6 atm, Let the maximum volume of the container when pressure, of CO2 is 1.6 atm be V L, During the process, PV = constant,  0.4 × 20 = 1.6 × V, 0.4  20, 5L,  V , 1.6, 17. (d) : Kp and Kc are related by the equation,, Kp = Kc(RT)ng, where ng = difference in the no. of moles of products, and reactants in the gaseous state., for 2C(s) + O2(g)  2CO2(g), ng = 2 – (1) = 1  0, 18. (a) : Fe(OH)3(s)  Fe3+ aq + 3OH–, K, , (, , [Fe3 ][OH ]3, , ), , (aq), , [Fe(OH)3 ], K = [Fe3+] [OH–]3, ( activity of solid is taken unity), Concentration of OH – ion in the reaction is decreased, by 1/4 times then equilibrium concentration of Fe3+ will, be increased by 64 times in order to keep the value of K, constant., 19. (a) : K p  pCO, 2, , Solids do not exert pressure, so their partial pressure is, taken as unity., 20. (a) : The value of K is high which means reaction, proceeds almost to completion i.e., the system will, contain mostly products., 21. (b) : 3H2 + N2  2NH3, 3, 3/2, 10 , 15, , 3, 2, , 1, 1/2, 10 , 5, , 2, 1, , 1, 2, , 10 × 1, 10, , Composition of gaseous mixture under the aforesaid, condition in the end will be, H2 = 30 – 15 = 15 litres, N2 = 30 – 5 = 25 litres ; NH3 = 10 litres, 22. (c) : N2(g), + 3H2(g)  2NH3(g), [NH ]2, 3, Kc , ; n(g) = 2 – 4 = –2, 3, [N2 ][H2 ], Thus, the reaction will go from right to left when Q > Kc., , www.neetujee.com, , 23. (a) : G = G° + RT lnQ, At equilibrium, G = 0 and Q = KC,  0 = G° + RT ln KC,  G° = –RT lnKC, = – 8.314 J K–1 mol–1 × 300 K × ln (2 × 1013), 24. (a), 25. (c) : When K > Q, rate of forward reaction > rate of, p, , backward reaction.,  Reaction is spontaneous., When G° < RT ln Q, G° is positive, reverse reaction is, feasible, thus reaction is non spontaneous., When Kp = Q, rate of forward reaction = rate of backward, reaction.,  Reaction is in equilibrium., When TS > H, G will be negative only when, H = +ve.,  Reaction is spontaneous and endothermic., 26. (a) : On increasing the pressure and decreasing the, temperature, equilibrium will shift in forward direction., 27. (d) : As the forward reaction is exothermic and, leads to lowering of pressure (produces lesser number, of gaseous moles) hence, according to Le Chatelier’s, principle, at high pressure and low temperature, the, given reversible reaction will shift in forward direction, to form more product., K p, H  1 1 , , 28. (a) : log, ,   , Kp, 2.303R T2 T1 , , , For exothermic reaction, H = –ve i.e., heat is evolved., The temperature T2 is higher than T1.,  1 1, Thus,    is negative.,  T2 T1 , so, log Kp – log Kp = –ve or log Kp > log Kp, or Kp > Kp, 29. (a) : HCl and SO2 are reducing agents. So, they can, reduce MnO–4., CO2 is neither oxidising nor reducing agent, it will, provide only acidic medium. It can shift the reaction in, forward direction and the reaction can go to completion., 30. (d) : X2(g )  4Y2(g ) 2XY4(g ), ng = –ve and H = –ve, The reaction is favoured in forward direction at low, temperature and high pressure., 31. (c) : CH4(g) + 2O2(g)  CO2(g) + 2H 2O(l), pCO2, K , p, 2, pCH  pO, 4, , 2, , www.mediit.in

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Equilibrium, , 65, , 32. (c) : Pressure of O2 does not depend on, concentration terms of other reactants (because both, are in solid state). Since this is an endothermic reaction,, if the temperature is raised, dissociation of BaO 2 would, occur, more O2 is produced at equilibrium, pressure of, O2 increases., 33. (b) : For a reaction, A + B  C + D,, Keq , , 44. (d) : Lewis acids are electron deficient compounds,, since (CH3)3B is electron deficient (due to incomplete, octet of B), it acts as a Lewis acid., 45. (d) : Due to strong hydrogen-fluorine bond, proton, is not given off easily and hence, HF is weakest acid., 46. (a) : B(OH)3 in aqueous medium coordinates, a molecule of water to form the hydrated species, , [C][D], , . In this species, B3+ ion, because of its, , [A][B], Increase in conc. of reactants will proceed the equilibrium, in the forward direction giving more products so that, the equilibrium constant value remains constant and, independent of concentration., 34. (d) : When solid and liquid are in equilibrium, the, increase in temperature results in increase in volume of, liquid or decrease in the amount of solid., Solid  Liquid, With increase in temperature equilibrium shifts in, forward direction., 35. (d) : According to Le Chatelier’s principle, if an, equilibrium is subjected to a change in concentration,, pressure or temperature, etc. equilibrium shift in such a, way so as to undo the effect of a change imposed., 36. (a) : HCl is a strong acid and dissociates completely, into ions in aqueous solution., 37. (b) : CH3COOH + H 2O  CH 3COO – + H3O +, As acetic acid is a weak acid so, it also contains some, undissociated CH3COOH along with CH3COO– and, H3O+ ions., 38. (d) : Bronsted acid Conjugate base, H 2O, OH–, HF, , F–, , 39. (c) : HCl cannot accept H+ ion, therefore cannot act, as Bronsted Base., 40. (b) : BF3, Lewis acid (incomplete octet), PF3, Lewis base (presence of lone pair on P atom), CF4, Complete octet, SiF4, Lewis acid (empty d-orbital in Si-atom), 41. (a) : BF3 is Lewis acid (e– pair acceptor)., +7, , 42. (a) : H Cl O4 with highest oxidation number and, its conjugate base is resonance stabilised, hence it is most, acidic. Cl is more electronegative than S., 43. (c) : Compounds that are electron deficient act as, Lewis acids. Out of the given hydrides B2H6 satisfies this, condition and is therefore a Lewis acid., , www.neetujee.com, , small size, has high polarizing power thereby pulling, the sigma electron charge of the coordinated O atom, towards itself. The coordinated oxygen, in turn, pulls the, sigma electron charge of the OH bond of the attached, water molecule towards itself. This facilitates the removal, of H+ ion from the O – H bond., Thus, the solution of B(OH)3 in water acts as a weak acid,, and it is not a protonic acid., 47. (c) : Proton accepting tendency is known as the, strength of basicity., In R—NH2, N has lone pair of electrons which intensify, due to electron releasing R-group and increase the, tendency to donate lone pair of electrons to H+., Secondly as the size of the ion increases there is less, attraction for H+ to form bond with H–atom and are less, basic. Thus the order of proton accepting tendency :, RNH2 > NH3 > HS– > I–., 48. (d) : NH – + H+  NH (conjugate acid), 2, , 3, , +, , Substance + H  conjugate acid, Substance – H+  conjugate base, 49. (b) : In BCl3 the central atom ‘B’ is sp2 hybridised, and contains only ‘six’ electrons in its valence shell., Therefore, it is electron deficient., 50. (d) : CH3COOH, Weak acid, , CH3 COO – + H+, Strong conjugate base, , As CH3COOH is the weakest acid, so its conjugate base, (CH3COO–) is the strongest base. H2SO4, HCl, HNO3, are strong acids, so their conjugate bases are weak., 51. (b) : In BF3 and FeCl3 molecules, the central atoms, have incomplete octet and in SiF4, the central atom has, empty d-orbitals. Hence, according to Lewis concept,, these are Lewis acids., 52. (a) : Ammonium sulphate is a salt of strong, acid (H2SO4) and weak base (NH4OH). Therefore,, repeated use of ammonium sulphate would increase the, concentration of sulphuric acid, while ammonia from, NH4OH is used up by the plant. Hence, the acidity of soil, will increase., , www.mediit.in

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66, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 53. (a) : Ni(OH)2 Ni2+  2OH, s, , s, , 58. (c) : Na2CO3 which is a salt of NaOH (strong base), and H2CO3 (weak acid) will produce a basic solution, with pH greater than 7., 59. (b) : Degree of dissociation,  3.7  0.037, 100, According to Ostwald’s formula,, Ka= 2C = (0.037)2 × 0.10 = 1.369 × 10–4  1.4 × 10–4, , 2s, , where s is the solubility of Ni(OH)2., NaOH Na+  OH, 0.1 M, , 0.1 M, , 0.1 M, , [OH–] = 2s + 0.1 ≈ 0.1 ( 2s  0.1), Ionic product of Ni(OH) = [Ni2+][OH –] 2, , s, , –14, , 2, , 2 × 10–15 =, s(0.1)2, 15, , 60. (d) : We know that,–14, at 25°C, Kw = 1 × 10 +, At 100°C, Kw = 55 × 10, [ Kw = [H ][OH–], , 2  10,  2  1013 M, 0.1  0.1, ,  Kw = [H +] 2, H  Kw, , 54. (c) : NaOH  Na+ + OH–, 0.01 M, , , , 0.01 M, , H  55 1014, pH = – log [H +], ,  [OH–] = 0.01 M, , , , pOH = – log [OH–] = – log(0.01) = 2, pH = 14 – pOH = 14 – 2 = 12, , 55. (d) : pH = 1, so [H+] = 10–1, For acid base mixture: N1V1 – N2V2 = N3V3, (For NaOH and HCl, Normality = Molarity), 1, 1, 60   40 , A. M1(H ) 10 10 = 2 × 10–2 M, 100, i.e. pH = 1.698  1.7, 1, 1, 55   45 , 1, B. M (H ) 10 10 , 2, , 100, , 100, = 10–2 M i.e. pH = 2, , 1, 1, 75   25 , C. M (H ) 5 5  101M i.e. pH  1, 3, 100, 1, 1, 100  100 ,  10 10,  0 i.e. pH  7, D. M4 (H ) , 200, , 56. (b): C5H5N+ H2O C5H5 NH  OH, , 61. (a) : BaCl2 is made up of Ba(OH)2 and HCl., AlCl3 is made up of Al(OH)3 and HCl., LiCl is made up of LiOH and HCl., BeCl2 is made up of Be(OH)2 and HCl., Ba(OH)2 is strongest base among the given options thus, have maximum pH., 62. (d) : CH3COOH  CH 3COO – + H +, C–x, , x, , CH3COONa  CH3COO, 0.2 M, , , , 0.10 M, , Kb, 1.7 109,  1.30 104, , C, 0.10,  Percentage of pyridine that forms pyridinium ion, = 1.30 × 10–4 × 100 = 0.013%, 57. (d) : One mole of NaOH is completely neutralised, , , by one mole of HCl., Hence, 0.01 mole of NaOH will be completely neutralised, by 0.01 mole of HCl.,  NaOH left unneutralised = 0.1 – 0.01 = 0.09 mol, As equal volumes of two solutions are mixed,, 0.09, , [OH] ,  0.045 M, 2,  pOH = –log(0.045) = 1.35,  pH = 14 – 1.35 = 12.65, , www.neetujee.com, , pH  log  55 1014 , 1, 1,  [ log(55  1014 )] [ log 55 14 log10], 2, 12, 1,  [1.74  14]  [12.26]  6.13, 2, 2, , 0.2 M, , –, , x, , + Na, , +, , 0.2 M, , acetic acid is a, weak acid so,, [CH3COOH] = C – x  0.1 M dissociation is, , [CH3COO–] = 0.2 + x  0.2 M , minimum i.e., x, can be neglected., K [CH3COOH],  [H+]  a, [CH3 COO  ], , , 1.8  105  101, 2  10, , 1, ,  9  106 M, , 63. (d) : NH4Cl is a salt of strong acid and weak base, so, hydrolysis constant is, K, Kh w, Kb, Given, Kb (NH4OH) = 1.77 × 10–5, Kw = 10–14,  K , , 1014, ,  0.565  109, , h, , or, , 1.77  105, Kh = 5.65 × 10–10, , www.mediit.in

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Equilibrium, , 67, , 64. (d) : Millimoles of H+ produced = 20 × 0.05 = 1, Millimoles of OH– produced = 30 × 0.1 × 2 = 6, ( Each Ba(OH)2 gives 2OH–.), –, , Millimoles of OH remaining in solution = 6 – 1 = 5, Total volume of solution = 20 + 30 = 50 mL, 5,  [OH ]   0.1 M, 50, 65. (d) +: pH = –log[H+], –3, or [H ] = 10–pH ; [H+] of soln. 1 = 10, [H+] of soln. 2 = 10–4 ; [H+] of soln. 3 = 10–5, Total concentration of [H+], –3, , –1, , –2, , = 10 (1 + 1 × 10 + 1 × 10 ), 1 , 1 1, 3 100  10  1,  10      10 , , 1 10 100 , 100, , , 111, , ,  103,  1.11  103, , , 100 , So, H+ ion concentration in mixture of equal volumes of, 1.11  103, 3, , these acid solution =, = 3.7 × 10–4 M, 3, 66. (a) : For a weak acid, degree of dissociation,, 2, Ka, 1 105,  10, , , i.e. 1.00%, C, 0.1, 67. (a) : Given, [H O+] = 1 × 10–10 or, pH = 10, 3, , Now at 25°C, pH + pOH = pKw = 14, or, pOH = 14 – pH = 14 – 10 = 4, 68. (c) : 10–8 M HCl = 10–8 M H+, Also from water, [H+] = 10–7, Total [H+] = 10–7 + 0.10 × 10–7 = 1.1 × 10–7 M, 69. (d) : C = 0.01 M, Kb = 1 × 10–12 at 25°C, BOH, B+ + OH –, Initially C, At eq. C – C, [OH–] = C, , 0, , 0, , C, , C, , , , 12, 2, [OH ]  Kb C  1 10  10, [OH–] = 10–7 mol L–1, , 70. (b) : NH4OH is a weak base but HCl is a strong acid, in solution, so pH of NH4Cl solution is comparatively, low., NaNO3 is a salt of strong base and strong acid, so pH of, the solution will be 7., Hydrolysis of potassium acetate (a salt of a weak acid and, a strong alkali) gives a weakly alkaline solution, since the, acetate ion acts as a weak base., CH3 COOK + H 2O  CH3 COOH + K + + OH–, The pH of this solution  8.8., , www.neetujee.com, , Hydrolysis of sodium carbonate (a salt of strong alkali, and a weak acid) gives an alkaline solution., Na2CO3 + 2H2O  2(Na+ + OH–) + H2CO3, The pH of this solution is > 10., 71. (d) : CH3COOH  CH 3COO – + H +, [CH3COO ][H+ ], K, a, [CH 3 COOH] 4, 3.4 10  3.4 104, = 6.8 × 10–3, [CH3COOH] =, 1.7 105, Ka, , , , 1, , , , 72. (b) : (i) H2 A HA  H, , , (ii) HA, , Ka, , 2, , 2, ,  , , , A, , H, , , , In the 1st step, H+ ion comes from neutral molecule, while, in the 2nd step the H+ ion comes from negatively charged, ions. The presence of –ve charge makes the removal of, H+ ion difficult. Thus, K > K ., a1, a2, 73. (b) :, After mixing 1 N solution of CH3COOH (weak, acid) and 1 N NaOH (strong base), the resulting solution, will have free OH– ions. Thus, pH will be higher than 7., 74. (a) : [H+] = C  = 0.1 × 0.02 = 2 × 10–3 M, (As degree of dissociation = 2% = 0.02), –12, 1014, –, Hence, [OH ] =, = 5 × 10 M, 2 103, 75. (c) : NaH + H2O  NaOH + H2, or, H–(aq) + H2O(l )  OH– + H2, 76. (d) : At high temperature, the value of ionic product, increases., 77. (c) :, , 2HI , , H2 + I2, , Initially :, After dissociation :, , 2, 2 – 2, , 0, , , 0, , , Total number of moles = 2(1 – ) + 2 = 2, 78. (b) : Since NaOH is a strong base, therefore it, completely ionises. Thus, the hydroxyl ion concentration, is equal to that of the base itself. We know that, concentration of OH– in N/10 NaOH = 0.1 = 10–1., Therefore value of, 1 1014, K,  1  1013, [H3O ]  w , [OH ], 101, pH = – log [H3O+] = – log [1 × 10–13] = 13, 79. (c) : Since HCl is a strong acid and it completely, ionises, therefore H3O+ ions concentration is equal that, of the acid itself i.e., [H3O+] = [HCl] = 10 M., Therefore, pH = – log [H3O+] = – log [10] = – 1, 80. (a) : [H3O+] = [OH–] = 1 × 10–6 mole/litre, Kw = [H 3O+][OH– ] = [1× 10–6] × [1 × 10–6] = 1 × 10 –12, , www.mediit.in

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68, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 81. (a) : Sodium borate is a salt formed from strong, base (NaOH) and weak acid (H3BO3). Hence, sodium, borate will act as basic solution., 82. (c) : NH4Cl and NaHCO3 are acidic in nature and, NaCl is neutral. Only Na2CO3 is basic and thus, have, highest pH., 83. (d) : Acid-base titration :, HCl, + NH4OH, NH4Cl, 10 mmol, , 20 mmol, , [H+][In ], KIn  [HIn], , ... (ii), , Kw = [H+] [OH–], From (ii) and (iii),, , ... (iii), , Kw [HIn][OH ],  Kh, KIn , [In ], K, [OH ]  w [In ], , ... (iv), , KIn [HIn], ,  HCl is the limiting reagent., Solution contains NH4OH (weak base) and NH4Cl (salt, of strong acid and weak base). Therefore, a basic buffer, will be formed., , [In ], log [OH–] = logKw – logKIn + log [HIn], [In ], –pOH = –pKw + pKIn + log, [HIn], , 84. (d) : Acidic buffer is a mixture of a weak acid and its, salt with a strong base. HClO4 is a strong acid., [Salt], 85. (b) : pH = pKa + log, [Salt] [Acid], 5 = – log Ka + log, [ pKa = – log Ka], [Acid], [Salt], 5 = – log [1 × 10–4] + log, [Acid], [Salt], [Salt], , 5  4  log, 5  4  log, [Acid], [Acid], [Salt] , [Salt], 1  log,  10 i.e. 10 :1, , [In ], pKw – pOH = pKIn + log, [HIn], [In ], or, pH = pK +, log, In, [HIn], [In ], i.e. log, = pH – pK, In, [HIn], 91. (b) : Solution of 0.1 N NH OH and 0.1 N NH Cl is, , [Acid] [Acid], 86. (a), 87. (b) : [NH3] = 0.30 M, Kb = 1.8 × 10–5, [NH +] = 0.20 M, 4, 5) = 4.74, pKb = –log(1.8 × 10, [salt], 0.2, pOH  pKb  log,  4.74  log,  4.56, [base], 0.3, pH = (14 – 4.56) = 9.44, [B ], 88. (d) : We know, pOH  pKb  log, [HB], Since, [B–] = [HB] (given),  pOH = pKb  pOH = 10,  pH = 14 – 10 = 4, 89. (b) : HNO2 (weak acid) and NaNO2 (salt of, conjugate base) is an example of acidic buffer., 90. (d) : Let us consider the formation of a salt of a weak, acid and a strong base., In– + H2O HIn + OH –, [HIn][OH  ], Kh , ...(i), [In ], Other equations present in the solution are, HIn H++ + In– –, H O H + OH, 2, , www.neetujee.com, , , , 4, , 4, , a buffer solution., According to Henderson equation, the pH of a basic, [Salt], buffer, pH = 14 – pKb – log, [Base], [Salt],  pK =14  pH  log, b, [Base], 0.1,  pKb  14  9.25  log, 0.1,  pKb = 14 – 9.25 = 4.75,  pKb of NH4OH = 4.75, 92. (b) : pH of an acidic buffer solution is given by, Henderson equation :, [Salt], pH  pK a log, [Acid], Its buffer capacity = pKa ± 1, Since a buffer solution is more effective in pH range, pKa ± 1 therefore, the weak acid having pKa = 3.58, together with its sodium salt is chosen. Acetoacetic acid, is, therefore, the suitable weak acid., 93. (d) : The pH value of the blood is maintained, constant by buffer solution present in the blood itself., Buffer solutions resist the change in pH values., 94. (b) : pH of the saturated solution of Ca(OH) = 9, 2, ,  pOH of the saturated solution of Ca(OH)2 = 14 – 9 = 5,  OH– = 10–5, ( pH + pOH = 14), Ca(OH)2  Ca2+ + 2OH–, s, , 1/2 × 10, , –5, , 2s –5, 10, , www.mediit.in

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Equilibrium, , 69, , Ksp = [Ca2+] [OH–]2 = [1/2 × 10–5] [10–5]2, = 0.5 × 10–15, , 100. (b) :, Salt, , 95. (c) : CaF2  Ca2+ + 2F–, , Ag CrO 1.1 × 10–12 = 4s3, , s, , 2s, , 2, , NaF  Na + F, +, , [Ca2+] = s, [F– ] = (2s + 0.1) ≈ 0.1 M, Ksp = [Ca2+] [F–]2, 5.3 × 10–11 = (s) (0.1)2, 5.3 1011, = 5.3 × 10–9 mol L–1, s=, –9, –1, (0.1)2,  Molar, solubility is 5.3 × 10 mol L, 96. (a) : Solubility of BaSO4,, 2.42 103, s, mol L1 = 1.04 × 10–5 mol L–1, 233, BaSO4 ionizes completely in the solution as :,  Ba2,  SO2, BaSO, 4(s), , (aq), s, , 4(aq), , AgCl, , 1.8 × 10–10 = s2, , AgBr, , 5 × 10–13 = s2, , s+ 2, , K, , s, , 2s, , = [Ag ] [C2O42–], Ksp = (2s)2(s) = 4s3, Ksp = 4 × (1.1 × 10–4)3, ([Ag+] = 2s = 2.2 × 10–4), Ksp = 5.3 × 10–12, 98. (b) : Let s be the solubility of AgCl in moles per litre., ,  , ,  Ag   Cl, AgCl, sp, , (aq), , (aq), , s, , s, , (aq), (s  0.1), , ( 0.1 M NaCl solution also provides 0.1 M Cl– ion)., Ksp = [Ag+] [Cl–]; 1.6 × 10–10 = s(s + 0.1), 1.6 × 10–10 =10, s(0.1), ( s < < < < 0.1), 1.6 10,  1.6 109 M, , s, , 0.1, 99. (d) : For MY : Ksp = s 12, s1  Ksp  6.2 1013 = 7.87 × 10–7 mol L–1, For NY : K = (s )(3s )3 = 27s 4, , , , 3, , sp, , 2, , 2, , 2, , 1013, s2, = 3.89 × 10–4 mol L–1, 27, Hence, molar solubility of MY in water is less than that, of NY3., , , 6.2, 4, , www.neetujee.com, , s  Ksp  0.7110, , 6, , be, , 63.3 × 103 J = – 2.303 × 8.314 × 298 logKsp, 63.3 × 103 J = –5705.84 logKsp, 63.3 103,  11.09, 5705.84, Ksp = antilog (–11.09) = 8.128 × 10–12, 102. (d) : CaCO3  Ca2+ + CO 2–, log Ksp  , , CaC2O4  Ca, , 97. (c) : Let solubility of Ag2C 2O4 be s mol L–1, +, 2–, Ag2 C2 O4(s)  2Ag(aq, ) + C 2O 4(aq, ), , s  Ksp  1.34 105, , precipitated at last., 101. (b) : G° = –2.303RT logKsp, , Ksp = [Ba2+] [SO4 ] = s2, = (1.04 × 10–5)2 = 1.08 × 10–10 mol2 L–2, , 4, , s  3 Ksp  0.65 10, 4, , 8, s  Ksp  0.9 10, AgI, 8.3 × 10–17 = s2, Solubility of Ag2CrO4 is highest thus, it will, , s, , 2–, , Solubility, , 4, , –, , 0.1 M 0.1 M, , Ksp, , 2+, , y, , x, , x, , 3, , 2–, , + C2O4, y, , Now, [Ca2+] = x + y, and x(x + y) = 4.7 × 10–9, y(x + y) = 1.3 × 10–9, Dividing equation (i) and (ii) we get, x,  3.6, y,  x = 3.6y, Putting this value in equation (ii), we get, y(3.6y + y) = 1.3 × 10–9, On solving, we get y = 1.68 × 10–5, and x = 3.6 × 1.68 × 10–5 = 6.05 × 10–5,  [Ca2+] = (x + y) = (1.68 × 10–5) + (6.05 × 10–5), , , ...(i), ...(ii), , [Ca2+] = 7.73 × 10–5 M, , 103. (b) : Sodium sulphide is soluble in water. The, solubility product (and hence solubility) of ZnS is larger, than that of CuS., 104. (b) : pH of solution = 12, [H+] = 10–12, 1014, [OH ] ,  102, 1012, 2+, ,  , Ba(OH) ,  Ba  2OH , 2, , s, , 102, , 2s, , 2s  102  s , , 2, Ksp = (s) (2s)2 = 4s3, , www.mediit.in

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70, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, ,  2 ,  4   10   4 106  5 107,  2  8, , 108. (b) : AgI  Ag+ +I–, KI  K+ + I–, , 105. (c) : Ksp[AgCl] = [Ag+][Cl–], , [For KI, 1 N = 1 M], [I–] = s + 10–4, Ksp = [Ag+][I–], 1 × 10–16 = s(s + 10–4), 1 × 10–16 = s2 + 10–4 s, , 3, , [Ag ] , , 1.8 1010, 1, , 10–4 M, ,  1.8 109 M, , 10, K [PbCl ] = [Pb2+][Cl–]2, 2, , sp, , 5, [Pb ]  1.7 10,  1.7 103 M, 2, , 101 101, 106. (a) : The cations of group II are precipitated as their, sulphides., Solubility product of sulphide of group II radicals2–are, very low. Therefore, even with low conc. of S ions,, the ionic product exceeds the value of their solubility, product and the radicals of group II gets precipitated., The low conc. of S2– ions is obtained by passing H2S, gas through the solution of the salt in the presence of, dil. HCl which suppresses degree of ionisation of H2S by, common ion effect., 2H+ + S 2–, , H2S, , H+, , + Cl, , common ion, Note that solubility product of group IV radicals are quite, high. It is necessary to suppress the conc. of S2– ions,, otherwise radical of group IV will also get precipitated, along with group II radicals., , 2, , s, 2, , 3, , 2s, , Ksp = s × (2s) = 4s ; i.e., 3.2 × 10, or, s3 = 0.8 × 10–11 = 8 × 10–12,  s = 2 × 10–4, , –11, , 3, , = 4s, , 1 × 10–16 = 10–4 s, [ s2 < < < 10–4 s], 16, 1  10,  1  1012 mol L1,  s , 4, 10, 109. (d) : Let s be the solubility of the electrolyte MX2., [M2+] = s, [X–] = 2s, Solubility product, Ksp = s × (2s)2 = 4s3;, s = 0.5 × 10–4 mol/litre, , , Ksp = 4 × (0.5 × 10–4)3; Ksp = 5 × 10–13, , 110. (b) : For reaction, M2S 2M + S, 2s, , +, , 2–, , s, , Solubility = 3.5 × 10–6, Solubility product, Ksp = [M+]2 [S2–], = (2s)2 s = 4s3 = 4 × (3.5 × 10–6)3 = 1.7 × 10–16, 111. (d) : For CaF2, decomposition is as follows :, CaF2  Ca2+ + 2F–, , –, , 107. (c) : Ksp = 3.2 × 10–11, AX A2  2X , , 10–4 M, , s, , 2s, ,  Ksp = [Ca2+] [F–]2 = s × (2s)2, or, , Ksp = 4s3  Ksp = 4 × (2 × 10–4)3, ,  Ksp = 32 × 10–12, 112. (b) : The greater the solubility product, the greater, is the solubility., 113. (a) : There are greater number of Cl– ions in CaCl2, compared to others. Hence, solubility of AgCl will be, minimum in 0.01M CaCl2 due to common ion effect., 114. (d) : Higher the value of solubility product, greater, is the solubility., , , , www.neetujee.com, , www.mediit.in

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72, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 10. (I) H2O2 + O3, (II) H2O2 + Ag2O, , (a) S 2O2–4 < SO2–3 < S 2O2–6, (b) SO2– < S O2– < S O2–, , H2O + 2O2, 2Ag + H2O + O2, , 3, , Role of hydrogen peroxide in the above reactions is, respectively, (a), (b), (c), (d), , oxidizing in (I) and reducing in (II), reducing in (I) and oxidizing in (II), reducing in (I) and (II), oxidizing in (I) and (II), , 4, , 26, , S2O2–6 <, 2–, , SO 2–, 3, < SO 2–, , (c), (d) S O < S O, 2, , 6, , 2, , 4, , (2003), , 3, , 17. Oxidation state of Fe in Fe3O4 is, (2014), , 11. The pair of compounds that can exist together is, (a) FeCl3, SnCl2, (b) HgCl2, SnCl2, (c) FeCl2, SnCl2, (d) FeCl3, KI, (2014), 12. A mixture of potassium chlorate, oxalic acid and, sulphuric acid is heated. During the reaction, which element undergoes maximum change in the, oxidation number?, (a) S, (b) H, (c) Cl, (d) C, (2012), 13. Oxidation numbers of P in PO 3–, of S in SO 2– and, 4, , 2, , S2O2–4 <, 2–, , 4, , 2–, that of Cr in Cr2O, are respectively, 7, (a) +3, +6 and +5, (b) +5, +3 and +6, , 4, 5, (b), 5, 4, 3, 8, (d), (c), (1999), 2, 3, 18. Reaction of sodium thiosulphate with iodine gives, (a) tetrathionate ion, (b) sulphide ion, (c) sulphate ion, (d) sulphite ion., (1996), (a), , 19. The oxide, which cannot act as a reducing agent is, (a) CO2, (b) ClO2, (c) NO2, (d) SO2, (1995), 20. Which substance is serving as a reducing agent in, the following reaction?, 14H+ + Cr O 2– + 3Ni  7H O + 2Cr 3+ + 3Ni2+, 2, , (a) H+, , 7, , 2, , (b) Cr O 2–, 27, , (c) –3, +6 and +6, (d) +5, +6 and +6, , (2009), , 14. Number of moles of MnO4– required to oxidize one, mole of ferrous oxalate completely in acidic medium, will be, (a) 7.5 moles, (b) 0.2 moles, (c) 0.6 moles, (d) 0.4 moles., (2008), 15. Which is the best description of the behaviour of, bromine in the reaction given below?, H2O + Br2  HOBr + HBr, (a) Proton acceptor only, (b) Both oxidised and reduced, (c) Oxidised only, (d) Reduced only, (2004), 16. The2–oxidation2–states of sulphur in the anions SO 2–,, 3, S O and S O follow the order, 2, , 4, , (1994), , (c) H2O, (d) Ni, 21. The oxidation state of I in H4IO – is, 6, , (b) – 1, (d) + 5, , (a) + 1, (c) + 7, , 8.4, , (1994), , Redox Reactions and Electrode Processes, , 22. Consider the change in oxidation state of bromine, corresponding to different emf values as shown in, the given diagram :, BrO– 1.82 V BrO– 1.5 V HBrO, 4, , 3, , Br– 1.0652 V Br2, , 1.595 V, , Then the– species undergoing disproportionation, is, (a) BrO, (b) BrO–, 4, (NEET 2018), (c) Br 3, (d) HBrO, 2, , 26, , ANSWER KEY, , 1., 11., 21., , (b), (c), (c), , www.neetujee.com, , 2., 12., 22., , (b), (c), (d), , 3., 13., , (c), (d), , 4., 14., , (b), (d), , 5., 15., , (b), (b), , 6., 16., , (c), (a), , 7., 17., , (a), (d), , 8., 18., , (b), (a), , 9., 19., , (d), (a), , 10., 20., , (c), (d), , www.mediit.in

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Redox Reactions, , 73, , Hints & Explanations, 1. (b) : Redox reactions are those, chemical, reactions which involve both oxidation and reduction, simultaneously., 2. (b) : Only ‘Al’ lies above ‘Zn’ in electrochemical, series, which can displace Zn from ZnCl2 solution., Therefore, conc. of ZnCl2 will decrease when kept in ‘Al’, container., 2Al + 3ZnCl2  2AlCl3 + 3Zn, 3. (c) : In CH 4, oxidation number of carbon is –4, while in CCl4, oxidation number of carbon is +4. Thus,, the change in oxidation number of carbon in the given, reaction is from –4 to +4., 4. (b), 5. (b) : Disproportionation reactions are those in, which the same element/compound gets oxidised and, reduced simultaneously., 2Cu+  Cu2+ + Cu0, +6, , +7, , +4, , 3MnO2–, + 4H +, 2MnO –4 + MnO 2 + 2H 2O, 4, 6. (c) : CrO5 has butterfly structure having two peroxo, bonds., Peroxo oxygen has –1 oxidation state., Let, state, Cr be, CrOoxidation, : x + 4(–1), + of, 1 (–2), = 0‘x’ x = +6, , O, , O, , O Cr, , O, O, , 5, , 7., 8., , +5 +2 0 –3, (a) : HNO3, NO, N2, NH4Cl, (b) : The correct balanced equation is, –, , 2–, , 2MnO4 + 5C2O4 + 16H+, , 2Mn2+ + 10CO2 + 8H2O, , 9. (d) : CaF2 + H2SO4  CaSO4 + 2HF, Here, the oxidation state of every atom remains the same, so, it is not a redox reaction., 10. (c) :, , H2O2 acts as reducing agent in both the reactions in, which O2 is evolved., 11. (c) : Both FeCl2 and SnCl2 are reducing agents with, low oxidation numbers., 12. (c) :, , www.neetujee.com, , Maximum change in oxidation number occurs in case of, chlorine, i.e., from +5 to –1., 13. (d) : Let oxidation number of P in PO3–4 be x.,  x + 4(–2) = –3  x = +5, Let oxidation number of S in SO4 2– be y.,  y + 4(–2) = –2  y = +6, 2–, Let oxidation number of Cr in Cr2O7 be z.,  2z + 7(–2), = –2  z = +6, 14. (d), : [5e– + MnO –4 + 8H+  Mn2+ + 4H O2 ..(i)] × 2, [C O 2–  2e–, + 2CO2.......(ii)] × 5, 2 4, On addition,, we, get, 2MnO– + 16H+ + 5C O2–  2Mn2+ + 10CO + 8H O, 4, , 2, , 4, , 2, , 2, , 2 moles of MnO–4 required to oxidise 5 moles of oxalate.,  Number of moles of MnO–4 required to oxidise, 1 mole of oxalate = 2/5 = 0.4, 15. (b) : H O  0    –1, Br2 HOBr HBr, 2, In the above reaction, the oxidation number of Br2, increases from zero (in Br2) to +1 (in HOBr) and, decreases from zero (in Br 2) to –1 (in HBr). Thus, Br 2, is oxidised as well as reduced and hence, it is a redox, reaction., 16. (a) : SO2–3 : x + (–2)3 = –2 or x – 6 = – 2 or x = + 4, S2O4 2– : 2x + (– 2)4 = – 2, , or 2x – 8 = –2 or 2x = +6  x = + 3, S2O6 2– : 2x + (– 2)6 = – 2, or 2x – 12 = – 2 or 2x = + 10  x = + 5, Oxidation states follow the order : S O2– < SO 2– < S O2–, 2 4, 26, 8 3, 17. (d) : Fe3O4 : 3x + 4(–2) = 0  x = +, 3, 18. (a) : 2Na2S2O3 + I2  Na2S4O6 + 2 NaI, (Sodium tetrathionate), , 19. (a) : Since carbon is in its maximum oxidation state, of +4, therefore, carbon dioxide (CO2) cannot act as a, reducing agent., 20. (d) : Since the oxidation number of Ni increases, from 0 to 2, therefore it acts as a reducing agent., 21. (c) : Let x = Oxidation state of I. Since oxidation, state of H = + 1 and oxidation state of O = – 2, therefore, for H4IO–6, we get, (4 × 1) + x + (6 × – 2) = – 1 or x = + 7, 22. (d) : For a reaction to be spontaneous, E°cell should, be positive as G° = –nFE°cell, HBrO, Br2 ; E°, = 1.595 V, SRP (cathode), HBrO, BrO– ; E° = –1.5 V, SOP (anode), 3, , 2HBrO, Br2 + BrO3–, E°cell = SRP (cathode) – SRP (anode), = 1.595 – 1.5 = 0.095 V, E°cell > 0  G° < 0 (spontaneous), , www.mediit.in

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, , , , , , , , , CHAPTER, , 9, , 9.1, 1., , 2., , 4., , 5., , The ionization of hydrogen atom would give rise to, (a) hydride ion, (b) hydronium ion, (c) proton, (d) hydroxyl ion., (1990), , 6., , 9.5, 7., , Water gas is produced by, (a) passing steam through a red hot coke, (b) saturating hydrogen with moisture, (c) mixing oxygen and hydrogen in the ratio of 1 : 2, (d) heating a mixture of CO2 and CH4 in petroleum, refineries., (1992), Which of the following metal evolves hydrogen on, reacting with cold dilute HNO3?, (a) Mg, (b) Al, (c) Fe, (d) Cu, (1989), , Properties of Dihydrogen, , Which of the following statements about hydrogen, is incorrect?, (a) Hydronium ion, H3O+ exists freely in solution., (b) Dihydrogen does not act as a reducing agent., , www.neetujee.com, , Hydrides, , Which of the following is electron-deficient?, (a) (BH3) 2, (b) PH3, (c) (CH3)2, (d) (SiH3)2 (NEET 2013), , 9.6, , Water, , 8., , The method used to remove temporary hardness of, water is, (a) synthetic resins method, (b) Calgon’s method, (c) Clark’s method, (d) ion-exchange method., (NEET 2019), , 9., , The number of hydrogen bonded water molecule(s), associated with CuSO4. 5H2O is, (a) 3, (b) 1, (c) 2, (d) 5, (Odisha NEET 2019), , 10., , Which of the following groups of ions makes the, water hard?, (a) Sodium and bicarbonate, (b) Magnesium and chloride, (c) Potassium and sulphate, (d) Ammonium and chloride, (1994), , 11., , At its melting point, ice is lighter than water because, (a) H2O molecules are more closely packed in solid, state, (b) ice crystals have hollow hexagonal arrangement, of H2O molecules, (c) on melting of ice the H2O molecules shrinks in, size, (d) ice forms mostly heavy water on first melting., (1992), , Preparation of Dihydrogen, H2, , Which one of the following pairs of substances on, reaction will not evolve H2 gas?, (a) Copper and HCl (aqueous), (b) Iron and steam, (c) Iron and H 2 SO4 (aqueous), (d) Sodium and ethyl alcohol, (1998), , 9.4, , (c) Hydrogen has three isotopes of which tritium is, the most common., (d) Hydrogen never acts as cation in ionic salts., (NEET-I 2016), , Position of Hydrogen in the Periodic, Table, , One would expect proton to have very large, (a) charge, (b) ionization potential, (c) hydration energy, (d) radius., (1993), , 9.3, 3., , Hydrogen, , www.mediit.in

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Hydrogen, , 75, –, , 9.7, 12., , (c) diatomic and form X ions, , Hydrogen Peroxide (H2O2), , 2, , (d) monoatomic and form X– ions., , Match the following and identify the correct option., (A) CO(g) + H2(g), (i) Mg(HCO3)2 +Ca(HCO3)2, (B) Temporary, (ii) An electron deficient, hardness of water, hydride, (C) B2H6, (iii) Synthesis gas, (D) H2O2, (iv) Non-planar structure, (A), (B), (C), (D), (a) (iii), (i), (ii), (iv), (b) (iii), (ii), (i), (iv), (c) (iii), (iv), (ii), (i), (d) (i), (iii), (ii), (iv), (NEET 2020), , 13. The structure of H2O2 is, (a) spherical, (b) non-planar, (c) planar, (d) linear., , 17. Which of the following is the true structure of H2O2?, H, , (a) H—O—O—H, , 14. The volume strength of 1.5 N H2O2 solution is, (a) 8.8, (b) 8.4, (c) 4.8, (d) 5.2, (1997,1996), 15. The O – O – H bond angle in H2O2 is, (a) 106°, (b) 109°28, (c) 120°, (d) 97°, , (1994), , 16. Hydrogen peroxide molecules2–are, (a) monoatomic and form X ions, , (b) O, , O, H, , H, , (c), , O, , (1989), , (d), , O, , H, , 18. The reaction of H2O2 with H2S is an example of, ......... reaction., (a) addition, (b) oxidation, (c) reduction, (d) acidic, (1988), , 9.8, (2003), , (1991), , Heavy Water, D2O, , 19. Some statements about heavy water are given below., (i) Heavy water is used as a moderator in nuclear, reactors., (ii) Heavy water is more associated than ordinary, water., (iii) Heavy water is more effective solvent than, ordinary water., Which of the above statements are correct?, (a) (i) and (ii), (b) (i), (ii) and (iii), (c) (ii) and (iii), (d) (i) and (iii), , 2, , (b) diatomic and form X– ions, , (Mains 2010), ANSWER KEY, , 1., 11., , (c), (b), , 2., 12., , (c), (a), , 3., 13., , (a), (b), , 4., 14., , (a), (b), , 5., 15., , (a), (d), , 6., 16., , (b,c) 7., (b) 17., , (a), (b), , 8., 18., , (c), (b), , 9., 19., , (b), (a), , 10., , (b), , Hints & Explanations, 1. (c) : Proton (H+) ion being very small in size would, have very large hydration energy., 2., , (c) : It gives rise to proton., +, –, H(g)  H(aq, )+ e, Proton, , 3. (a) : Copper is a noble metal, as it lies below, hydrogen in the electrochemical series. Therefore, it, cannot displace hydrogen from dilute HCl. While iron, and sodium lie above hydrogen in the electrochemical, series, so they can liberate H2 either from steam or H2SO4, solution., C2H5 – OH + Na  C2H5 – ONa + 1/2H2, , www.neetujee.com, , Fe + H2SO4  FeSO4 + H2, 3Fe + 4H2O  Fe3O4 + 4H2, 4. (a) : H2O  C  H2  CO, Steam, , Red hot, , Water gas, , 5. (a) : Mg reacts with nitric acid to give Mg(NO3)2, and evolves H2 gas., Mg + 2HNO3  Mg(NO3)2 + H2, 6. (b, c) : Dihydrogen acts as a powerful reducing, agent and reduces metal oxides such as CuO, ZnO, PbO, and Fe3O4 to their respective metals., CuO + H2  Cu + H2O, ZnO + H2  Zn + H2O, , www.mediit.in

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76, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , Fe3O4 + 4H2  3Fe + 4H2O, Hydrogen has three isotopes of which protium is the, most common and tritium is radioactive., 7. (a) : Boron hydrides are electron-deficient, compounds., 8. (c) : Clark’s process is used to remove temporary, hardness of water. In this method quick lime is added., The bicarbonates present in temporary hard water, react with lime water to form insoluble calcium and, magnesium carbonates which can be easily filtered off., CaO + H2O, Ca(OH) 2, Quick lime, , Lime water, , Ca(HCO3)2 + Ca(OH)2, Mg(HCO3)2 + 2Ca(OH)2, , 2CaCO3 + 2H2O, 2CaCO3 +, Mg(OH)2 + 2H2O, , 9. (b) : The ionic formulation of CuSO4.5H2O is, [Cu(H2O)4]H2OSO4, in which four H2O molecules are, coordinated to a central Cu2+ ion while the fifth H 2O, molecule is hydrogen bonded to sulphate group., 10. (b) : Hardness of water is due to the presence of, bicarbonates, chlorides and sulphates of Ca and Mg., Hence, hard water will consist of Mg2+ and Cl– ions., 11. (b) : In ice crystals, water molecules are linked, through H-bonds in hollow hexagonal arrangement so,, volume is large and density is less. In liquid state this, hollow arrangement breaks into more closer arrangement, of molecules. Consequently the density is increased in, liquid state., 12. (a), , 13. (b) : Hydrogen peroxide has a non-planar structure., 14. (b) : Normality (N) = 1.5, We know that equivalent weight of H2O2 is 17 and, strength of H2O2 = Normality × Equivalent weight, = 1.5 × 17 = 25.5, 2H2O2,  2H2O + O2, (2 × 34 = 68 g), , (22.4 L), , Since 68 grams of H2O2 produces 22.4 litres oxygen at, NTP, therefore, 25.5 grams of H2O2 will produce, 22.4, × 25.5 = 8.4 litre of oxygen., , 68, Thus, volume strength of given H2O2 solution is 8.4., 15. (d) : Bond angle of O – O – H in H2O2 is 97°., 16. (b) : H O is diatomic and forms H+ + HO– (X–), 22, , 2, , (hydroperoxide ion)., H, , 17. (b) : O, , O is the true structure of H2O2., H, , 18. (b) : It is an example of oxidation reaction., H2S + H2O2  2H2O + S, H2O2 oxidises H2S into S., 19. (a) : Heavy water is used for slowing down the speed, of neutrons in nuclear reactors, hence used as moderator., Boiling point of heavy water is greater (374.42 K) than, that of ordinary water (373 K), hence heavy water is, more associated. Dielectric constant of ordinary water is, greater than that of heavy water, hence ordinary water is, a better solvent., , , , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 10, , The s-Block Elements, , 10.1 Group 1 Elements : Alkali Metals, 1., , Ionic mobility of which of the following alkali metal, ions is lowest when aqueous solution of their salts, are put under an electric field?, (a) K, (b) Rb, (c) Li, (d) Na (NEET 2017), , 2., , Which one of the alkali metals, forms only, the, normal oxide, M2O on heating in air?, (a) Rb, (b) K, (c) Li, (d) Na, (2012), , 3., , 4., , 5., , The ease of adsorption of the hydrated alkali metal, ions on an ion-exchange resins follows the order, (a) Li+ < K+ < Na+ < Rb+, (b) Rb+ < K+ < Na+ < Li+, (c) K+ < Na+ < Rb+ < Li+, (d) Na+ < Li+ < K+ < Rb+, (2012), The sequence of ionic mobility in aqueous solution, is, (a) Rb+ > K+ > Cs+ > Na+, (b) Na+ > K+ > Rb+ > Cs+, (c) K+ > Na+ > Rb+ > Cs+, (d) Cs+ > Rb+ > K+ > Na+, (2008), When a substance (A) reacts with water it produces, a combustible gas (B) and a solution of substance (C), in water. When another substance (D) reacts with, this solution of (C), it also produces the same gas (B), on warming but (D) can produce gas (B) on reaction, with dilute sulphuric acid at room temperature., Substance (A) imparts a deep golden yellow colour, to a smokeless flame of Bunsen burner. Then (A),, (B), (C) and (D) respectively are, (a) Ca, H2, Ca(OH)2, Sn, (b) K, H2, KOH, Al, (c) Na, H2, NaOH, Zn, (d) CaC2, C2H2, Ca(OH)2, Fe, (1998), , www.neetujee.com, , 6., , Which one of the following properties of alkali, metals increases in magnitude as the atomic number, rises?, (a) Ionic radius, (b) Melting point, (c) Electronegativity, (d) First ionization energy, (1989), , 10.2 General Characteristics of the, Compounds of the Alkali Metals, 7., , In the case of alkali metals, the covalent character, decreases in the order, (a) MF > MCl > MBr > MI, (b) MF > MCl > MI > MBr, (c) MI > MBr > MCl > MF, (d) MCl > MI > MBr > MF, (2009), , 8., , The alkali metals form salt-like hydrides by the, direct synthesis at elevated temperature. The thermal, stability of these hydrides decreases in which of the, following orders?, (a) NaH > LiH > KH > RbH > CsH, (b) LiH > NaH > KH > RbH > CsH, (c) CsH > RbH > KH > NaH > LiH, (d) KH > NaH > LiH > CsH > RbH, (2008), , 9., , Which compound will show the highest lattice, energy?, (a) RbF, (b) CsF, (c) NaF, (d) KF, (1993), , 10.3 Anomalous Properties of Lithium, 10., , Which of the alkali metal chloride (MCl) forms its, dihyrate salt (MCl.2H2O) easily ?, (a) LiCl, (b) CsCl, (c) RbCl, (d) KCl, (Odisha NEET 2019), , 10.4 Some Important Compounds of Sodium, 11., , Crude sodium chloride obtained by crystallisation, of brine solution does not contain, , www.mediit.in

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78, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (a) MgSO4, (c) MgCl2, 12., , 13., , 14., , (b) Na2SO4, (d) CaSO4, (Odisha NEET 2019), , In Castner-Kellner cell for production of sodium, hydroxide, (a) brine is electrolyzed using graphite electrodes, (b) molten sodium chloride is electrolysed, (c) sodium amalgam is formed at mercury cathode, (d) brine is electrolyzed with Pt electrodes., (Karnataka NEET 2013), Which of the following statements is incorrect?, (a) Pure sodium metal dissolves in liquid ammonia, to give blue solution., (b) NaOH reacts with glass to give sodium silicate., (c) Aluminium reacts with excess NaOH to give, Al(OH)3., (d) NaHCO3 on heating gives Na2CO3., (Mains 2011), In which of the following processes, fused sodium, hydroxide is electrolysed at a 330 °C temperature for, extraction of sodium?, (a) Castner’s process (b) Down’s process, (c) Cyanide process (d) Both (b) and (c)., (2000), , 15., , Which of the following is known as fusion mixture?, (a) Mixture of Na2CO3 + NaHCO3, (b) Na2CO3.10H2O, (c) Mixture of K2CO3 + Na2CO3, (d) NaHCO3, (1994), , 16., , Washing soda has formula, (a) Na2CO3.7H2O, (b) Na2CO3.10H2O, (c) Na2CO3.3H2O, (d) Na2CO3, (1990), , 10.5 Biological Importance of Sodium and, Potassium, 17., , 18., , The following metal ion activates many enzymes,, participates in the oxidation of glucose to produce, ATP and with Na, is responsible for the transmission, of nerve signals., (a) Iron, (b) Copper, (c) Calcium, (d) Potassium, (NEET 2020), The function of “Sodium pump” is a biological, process operating in each and every cell of all, animals. Which of the following biologically, important ions is also a constituent of this pump?, (a) K+, (b) Fe2+, (c) Ca2+, (d) Mg2+ (2015, Cancelled), , www.neetujee.com, , 10.6 Group 2 Elements : Alkaline Earth Metals, 19., , 20., , 21., , 22., , Magnesium reacts with an element (X) to form, an ionic compound. If the ground state electronic, configuration of (X) is 1s2 2s2 2p3, the simplest, formula for this compound is, (a) Mg2X3, (b) MgX2, (c) Mg2X, (d) Mg3X2 (NEET 2018), Electronic configuration of calcium atom may be, written as, (a) [Ne] 4p2, (b) [Ar] 4s2, 2, (c) [Ne] 4s, (d) [Ar] 4p2, (1992), Compared with the alkaline earth metals, the alkali, metals exhibit, (a) smaller ionic radii, (b) highest boiling points, (c) greater hardness, (d) lower ionization energies., (1990), Which of the following atoms will have the smallest, size?, (a) Mg, (b) Na, (c) Be, (d) Li, (1989), , 10.7 General Characteristics of Compounds, of Alkaline Earth Metals, 23., , 24., , 25., , 26., , 27., , HCl was passed through a solution of CaCl2, MgCl2, and NaCl. Which of the following compound(s), crystallise(s)?, (a) Both MgCl2 and CaCl2, (b) Only NaCl, (c) Only MgCl2, (d) NaCl, MgCl2 and CaCl2, (NEET 2020), Which of the following is an amphoteric hydroxide?, (a) Be(OH)2, (b) Sr(OH)2, (c) Ca(OH)2, (d) Mg(OH)2, (NEET 2019), Among CaH2, BeH 2, BaH2, the order of ionic, character is, (a) BeH2 < CaH2 < BaH2, (b) CaH2 < BeH2 < BaH2, (c) BeH2 < BaH2 < CaH2, (d) BaH2 < BeH2 < CaH2, (NEET 2018), On heating which of the following releases CO2, most easily?, (a) Na2CO3, (b) MgCO3, (c) CaCO3, (d) K2CO3, (2015), Solubility of the alkaline earth metal sulphates in, water decreases in the sequence, (a) Sr > Ca > Mg > Ba (b) Ba > Mg > Sr > Ca, (c) Mg > Ca > Sr > Ba (d) Ca > Sr > Ba > Mg, (2015, Cancelled), , www.mediit.in

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The s-Block Elements, 28., , 29., , 30., , 31., , 79, , Which of the following compounds has the lowest, melting point?, (a) CaCl2, (b) CaBr2, (c) CaI2, (d) CaF2, (2011), Which of the following alkaline earth metal, sulphates has hydration enthalpy higher than the, lattice enthalpy?, (a) CaSO4, (b) BeSO4, (c) BaSO4, (d) SrSO4, (2010), Which one of the following compounds is a, peroxide?, (a) KO2, (b) BaO2, (c) MnO2, (d) NO2, (2010), Property of the alkaline earth metals that increases, with their atomic number, (a) solubility of their hydroxides in water, (b) solubility of their sulphates in water, (c) ionization energy, (2010), , (d) electronegativity, , 32. Which of the following oxides is not expected to, react with sodium hydroxide?, (a) CaO, (b) SiO, 2, , (c) BeO, , (d) B2O3, , (2009), , 33. The correct order of increasing thermal stability of, K2CO3, MgCO3, CaCO3 and BeCO 3 is, (a) BeCO3 < MgCO3 < CaCO3 < K2CO3, (b) MgCO3 < BeCO3 < CaCO3 < K2CO3, (c) K2CO3 < MgCO 3 < CaCO3 < BeCO3, (d) BeCO3 < MgCO3 < K2CO3 < CaCO3, (2007), 34. In which of the following the hydration energy is, higher than the lattice energy?, (a) MgSO4, (b) RaSO4, (c) SrSO4, (d) BaSO4, (2007), 35. The solubility in water of sulphate down the Be, group is Be > Mg > Ca > Sr > Ba. This is due to, (a) decreasing lattice energy, (b) high heat of solvation for smaller ions like Be2+, (c) increase in melting points, (d) increasing molecular weight., (1995), 36. All the following substances react with water. The, pair that gives the same gaseous product is, (a) K and KO2, (b) Na and Na2O2, (c) Ca and CaH2, (d) Ba and BaO2. (1994), 37. Which of the following statement is false?, (a) Strontium decomposes water readily than, beryllium., (b) Barium carbonate melts at a higher temperature, than calcium carbonate., , www.neetujee.com, , (c) Barium hydroxide is more soluble in water than, magnesium hydroxide., (d) Beryllium hydroxide is more basic than barium, hydroxide., (1994), , 10.8 Anomalous Behaviour of Beryllium, 38. In context with beryllium, which one of the, following statements is incorrect?, (a) It is rendered passive by nitric acid., (b) It forms Be2C., (c) Its salts rarely hydrolyse., (d) Its hydride is electron-deficient and polymeric., (NEET-II 2016), , 10.9 Some Important Compounds of Calcium, 39. The suspension of slaked lime in water is known as, (a) lime water, (b) quick lime, (c) milk of lime, (d) aqueous solution of slaked lime. (NEET-II 2016), 40. The product obtained as a result of a reaction of, nitrogen with CaC is, 2, , (a) CaCN3, (b) Ca2CN, (c) Ca(CN)2, (d) CaCN (NEET-I 2016), 41. Which one of the following is present as an active, ingredient in bleaching powder for bleaching, action?, (a) CaOCl2, (b) Ca(OCl)2, (c) CaO2Cl, (d) CaCl2, (2011), 42. Match List-I with List-II for the compositions of, substances and select the correct answer using the, code given :, List-I, List-II, (Substances), (Composition), (A) Plaster of Paris, (i) CaSO4·2H2O, (B) Epsomite, (ii) CaSO4·1/2 H2O, (C) Kieserite, (iii) MgSO4·7H2O, (D) Gypsum, (iv) MgSO4·H2O, (v) CaSO4, (a) (A)-(iii), (B)-(iv), (C)-(i), (D)-(ii), (b) (A)-(ii), (B)-(iii), (C)-(iv), (D)-(i), (c) (A)-(i), (B)-(ii), (C)-(iii), (D)-(v), (d) (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i) (Mains 2011), 43. The compound A on heating gives a colourless gas, and a residue that is dissolved in water to obtain B., Excess of CO2 is bubbled through aqueous solution, of B, C is formed which is recovered in the solid, form. Solid C on gentle heating gives back A. The, compound is, (a) CaCO3, (b) Na2CO3, (c) K2CO3, (d) CaSO4·2H2O, (Mains 2010), , www.mediit.in

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80, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 44. Which of the, chlorite?, (a) Ca(ClO3)2, (c) CaClO2, , following represents, , calcium, , (b) Ca(ClO2)2, (d) Ca(ClO4)2, , (1996), , 45. Identify the correct statement., (a) Plaster of Paris can be obtained by hydration of, gypsum., (b) Plaster of Paris is obtained by partial oxidation, of gypsum., (c) Gypsum contains a lower percentage of calcium, than Plaster of Paris., (d) Gypsum is obtained by heating Plaster of Paris., (1995), 46. Bleaching powder is obtained by the action of, chlorine gas and, (a) dilute solution of Ca(OH)2, (b) concentrated solution of Ca(OH)2, (c) dry CaO, (d) dry slaked lime., (1988), , 10.10 Biological Importance of Magnesium, and Calcium, 47. Enzymes that utilize ATP in phosphate transfer, require an alkaline earth metal (M) as the cofactor., M is, (a) Sr, (b) Be, (c) Mg, (d) Ca, (NEET 2019), 48. Which of the following statements is false?, (a) Ca 2+ ions are not important in maintaining the, regular beating of the heart., (b) Mg2+ ions are important in the green parts of, the plants., (c) Mg2+ ions form a complex with ATP., (d) Ca2+ ions are important in blood clotting., (NEET-I 2016), 49. Which of the following metal ions play an important, role in muscle contraction?, (a) K+, (b) Na+, (c) Mg2+, (d) Ca2+, (1994), , ANSWER KEY, , 1., 11., 21., 31., 41., , (c), (a), (d), (a), (b), , 2., 12., 22., 32., 42., , (c), (c), (c), (a), (b), , 3., 13., 23., 33., 43., , (b), (c), (b), (a), (a), , 4., 14., 24., 34., 44., , (d), (a), (a), (a), (b), , 5., 15., 25., 35., 45., , (c), (c), (a), (b), (c), , 6., 16., 26., 36., 46., , (a), (b), (b), (c), (d), , 7., 17., 27., 37., 47., , 8., 18., 28., 38., 48., , (c), (d), (c), (d), (c), , (b), (a), (c), (c), (a), , 9., 19., 29., 39., 49., , (c), (d), (b), (c), (d), , 10., 20., 30., 40., , (a), (b), (b), (c), , Hints & Explanations, 1. (c) : The hydration enthalpy of alkali metal ions, decreases with increase in ionic sizes i.e.,, Li+ > Na+ > K+ > Rb+ > Cs+, Hence, lithium having maximum degree of hydration, will be least mobile., The order of ionic mobility is, [Li(aq) ] + < [Na(aq))] + < [K(aq) ] + < [Rb(aq)] +, 2. (c) : When alkali metals heated in atmosphere of, oxygen, the alkali metals ignite and form oxides. On, combustion Li forms Li2O; sodium gives the peroxide, Na2O2 and potassium and rubidium give superoxide, (MO2)., 3. (b) : The order of decreasing hydration enthalpy of, alkali metal ions is : Li+ > Na+ > K+ > Rb+, Thus, ease of adsorption of hydrated ions is in the order :, Rb+ < K+ < Na + < Li+., , www.neetujee.com, , 4. (d) : Smaller the size of cation, higher will be the, hydration and its effective size will increase and hence, mobility in aqueous solution will decrease. Hence, the, correct sequence of ionic mobility in aqueous solution of, the given cations is Cs+ > Rb+ > K+ > Na+., 5. (c) : Only ‘Na’ imparts golden colour to Bunsen, flame, therefore, A = Na, B = H2, C = NaOH, D = Zn., 2Na + 2H2O  2NaOH + H2, (A), , (C), , (B), , Zn + 2NaOH  Na2ZnO 2 + H2, (D), , (C), , (B), , Zn + H2SO4(dil.)  ZnSO4 + H2, (D), , (B), , 6. (a) : In a group, ionic radius increases with increase, in atomic number whereas the m.pt. decreases down in, a group due to weakening of metallic bonds. Similarly,, , www.mediit.in

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The s-Block Elements, , 81, , electronegativity and ionization energy also decrease, down the group., 7. (c) : Alkali metals are highly electropositive and, halogens are electronegative. Thus, for the halides of a, given alkali metal, the covalent character decreases with, increase in electronegativity of halogens.,  Order of covalent character of halides is, MI > MBr > MCl > MF., 8. (b) : The ionic character of the bonds in hydrides, increases from LiH to CsH due to weakening of M—H, bond so, thermal stability of these hydrides decreases in, the order of LiH > NaH > KH > RbH > CsH., 9. (c) : With the same anion, smaller the size of, the cation, higher is the lattice energy. Therefore, NaF, will show the highest lattice energy among the given, compounds., 10. (a) : LiCl is deliquescent and crystallises from, aqueous solution as hydrates, LiCl.2H2O., 11. (a) : Crude sodium chloride, generally obtained by, crystallisation of brine solution contains sodium sulphate, (Na2SO4), calcium sulphate (CaSO4), calcium chloride, (CaCl2) and magnesium chloride (MgCl 2) as impurities., Crude sodium chloride does not contain MgSO4., 12. (c) : In Castner-Kellner cell, sodium amalgam is, formed at mercury cathode., A brine solution is electrolysed using a mercury cathode, and a carbon anode., 13. (c) : Al reacts with NaOH to give sodium aluminate., 14. (a) : In Castner’s process, for production of sodium, metal, sodium hydroxide (NaOH) is electrolysed at, temperature 330 °C., 15. (c) : K2CO3 and Na2CO3 mixture is called as fusion, mixture., 16. (b) : Na2CO3.10H2O is washing soda., 17. (d) : Potassium ions are the most abundant cations, within cell fluids, where they activate many enzymes,, participate in the oxidation of glucose to produce ATP, and, with sodium, are responsible for the transmission of, nerve signals., 18. (a), 19. (d) : Electronic configuration of X is 1s2, 2s2 2p3., So, valency of X will 2+, be 3., Magnesium ion = Mg, 2+, , Formula : Mg3X2, , www.neetujee.com, , Mg, , X 3–, , 2, , 3, , 20. (b) : 20Ca, 18Ar, , 1s2, 2s22p6, 3s23p6, 4s2, 1s2, 2s22p6, 3s23p6, , Hence, 20Ca, [Ar]4s2, 21. (d) : The alkali metals are larger in size and have, smaller nuclear charge thus they have lower ionization, energy in comparison to alkaline earth metals., 22. (c) : The atomic size decreases within a period from, left to right, therefore Li > Be and Na > Mg. The size, increases in a group from top to bottom. Hence, the size, of Na is greater than Li. Overall order Na > Mg > Li > Be., Thus, Be has smallest size., 23. (b) : CaCl2 and MgCl2 are more soluble than NaCl., Thus, when HCl was passed through a solution containing, CaCl2, MgCl2 and NaCl, only NaCl got crystallised., 24. (a) : Be(OH)2 is amphoteric in nature as it reacts, with acid and alkali both., Be(OH)2 + 2OH–  [Be(OH)4]2–, Be(OH)2 + 2HCl + 2H2O  [Be(OH)4]Cl2, 25. (a) : BeH2 < CaH2 < BaH2, On moving down the group, metallic character of metals, increases. So, ionic character of metal hydrides increases., Hence, BeH2 will be least ionic., 26. (b) : Stability of carbonates increases down the, group with increase in the size of metal ion. Also the, alkali metal carbonates are more stable than alkaline, earth metal carbonates., Hence, MgCO3 is least stable and it releases CO2 most, easily., , ,  MgO + CO2, MgCO3 , 27. (c) : Solubility of alkaline earth metal sulphates, decreases down the group because hydration energy, decreases., 28. (c) : As the covalent character in compound, increases and ionic character decreases, melting point, of the compound decreases. So, CaI2 has the highest, covalent character and lowest melting point., 29. (b) : The hydration enthalpy of BeSO 4 is higher than, its lattice energy. Within group 2, the hydration energy, decreases down the group while lattice energy is almost, the same., 30. (b) : BaO2 has peroxide linkage., 31. (a) : The solubility of an ionic compound depends, on two factors :, (a) lattice energy, and (b) hydration energy, In case of alkaline earth metal hydroxides, the lattice, energy decreases as we move down the group. This, decrease is more than the decrease in the hydration, energy down the group., , www.mediit.in

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, , , , , , , , , CHAPTER, , 11, , The p-Block Elements, , 11.1 Group 13 Elements : The Boron Family, 1., , The correct order of atomic radii in group 13, elements is, (a) B < Al < In < Ga < Tl, (b) B < Al < Ga < In < Tl, (c) B < Ga < Al < Tl < In, (d) B < Ga < Al < In < Tl, (NEET 2018), , 2., , AlF3 is soluble in HF only in presence of KF. It is due, to the formation of, (a) K3[AlF3H3], (b) K3[AlF6], (c) AlH3, (d) K[AlF3H], (NEET-II 2016), , 3., , The stability of +1 oxidation state among Al, Ga, In, and Tl increases in the sequence, (a) Al < Ga < In < Tl (b) Tl < In < Ga < Al, (c) In < Tl < Ga < Al (d) Ga < In < Al < Tl, (2015, 2009), , 4., , Aluminium(III) chloride forms a dimer because, aluminium, (a) belongs to 3rd group, (b) can have higher coordination number, (c) cannot form a trimer, (d) has high ionization energy., (1995), , 11.2 Important Trends, Properties of Boron, 5., , 6., , and, , Anomalous, , Which one of the following elements is unable to, form MF63– ion?, (b) Al, (a) Ga, (c) B, (d) In, (NEET 2018), The tendency of BF3, BCl3 and BBr3 to behave as, Lewis acid decreases in the sequence, (a) BCl3 > BF3 > BBr3, (b) BBr3 > BCl3 > BF3, (c) BBr3 > BF3 > BCl3, (d) BF3 > BCl3 > BBr3, (2010), , www.neetujee.com, , 7., , Boron compounds behave as Lewis acids, because of, their, (a) ionisation property, (b) electron deficient nature, (c) acidic nature, (d) covalent nature., (1996), , 11.3 Some Important Compounds of Boron, 8., , Boric acid is an acid because its molecule, (a) contains replaceable H+ ion, (b) gives up a proton, (c) accepts OH– from water releasing proton, (d) combines with proton from water, molecule., (NEET-II 2016), , 9., , Which of the following structure is similar to graphite?, (a) B4C, (b) B2H6, (c) BN, (d) B, (NEET 2013), , 10. The type of hybridisation of boron in diborane is, (a) sp3-hybridisation, (b) sp2-hybridisation, (c) sp-hybridisation, (d) sp3d2-hybridisation., (1999), 11. Which of the following statements about H3BO 3 is, not correct?, (a) It has a layer structure in which planar BO3, units are joined by hydrogen bonds., (b) It does not act as proton donor but acts as a, Lewis acid by accepting hydroxyl ion., (c) It is a strong tribasic acid., (d) It is prepared by acidifying an aqueous solution, of borax., (1994), , 11.5 Group 14 Elements : The Carbon Family, 12. Which of the following is incorrect statement ?, (a) SnF4 is ionic in nature., (b) PbF4 is covalent in nature., (c) SiCl4 is easily hydrolysed., (d) GeX4 (X = F, Cl, Br, I) is more stable than GeX2., (NEET 2019), , www.mediit.in

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84, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 13. Which of the following species is not stable?, (a) [SiCl6]2–, (b) [SiF6]2–, 2–, (c) [GeCl6], (d) [Sn(OH) 6] 2–, (NEET 2019), 14. It is because of inability of ns2 electrons of the, valence shell to participate in bonding that, (a) Sn2+ is oxidising while Pb 4+ is reducing, (b) Sn2+ and Pb2+ are both oxidising and reducing, (c) Sn4+ is reducing while Pb4+ is oxidising, (d) Sn2+ is reducing while Pb4+ is oxidising., (NEET 2017), 15. Which of the following oxidation states are the most, characteristic for lead and tin respectively?, (a) +2, +4, (b) +4, +4, (c) +2, +2, (d) +4, +2, (2007), 16. Carbon and silicon belong to (IV) group. The, maximum coordination number of carbon in, commonly occurring compounds is 4, whereas that, of silicon is 6. This is due to, (a) availability of low lying d-orbitals in silicon, (b) large size of silicon, (c) more electropositive nature of silicon, (d) both (b) and (c)., (1994), , 11.7 Allotropes of Carbon, , (1999), , 19. In graphite, electrons are, (a) localised on each C-atom, (b) localised on every third C-atom, (c) spread out between the structure, (d) present in antibonding orbital., (1997, 1993), 20. Which of the following types of forces bind together, the carbon atoms in diamond?, (a) Ionic, (b) Covalent, (c) Dipolar, (d) van der Waals (1992), 21. Which of the following is an insulator?, (a) Graphite, (b) Aluminium, (c) Diamond, (d) Silicon, , www.neetujee.com, , 22. Identify the correct statements from the following :, (A) CO2(g) is used as refrigerant for ice-cream and, frozen food., (B) The structure of C60 contains twelve six carbon, rings and twenty five carbon rings., (C) ZSM-5, a type of zeolite, is used to convert, alcohols into gasoline., (D) CO is colourless and odourless gas., (a) (A), (B) and (C) only, (b) (A) and (C) only, (c) (B) and (C) only, (d) (C) and (D) only, (NEET 2020), 23. Which of the following compounds is used in, cosmetic surgery?, (a) Silica, (b) Silicates, (c) Silicones, (d) Zeolites, (Odisha NEET 2019), 24. Which of these is not a monomer for a high, molecular mass silicone polymer?, (a) Me3SiCl, (b) PhSiCl3, (c) MeSiCl3, (d) Me2SiCl2, (NEET 2013), 25. The basic2–structural unit of silicates, is, (a) SiO3, (b) SiO42–, (c) SiO–, , 17. Which of the following does not show electrical, conduction?, (a) Diamond, (b) Graphite, (c) Potassium, (d) Sodium, (1999), 18. Percentage of lead in lead pencil is, (a) 80, (b) 20, (c) zero, (d) 70, , 11.8 Some Important Compounds of Carbon, and Silicon, , (1992), , (d) SiO4 4–, , (NEET 2013), , 26. Which statement is wrong?, (a) Beryl is an example of cyclic silicate., (b) Mg2SiO4 is orthosilicate., (c) Basic structural unit in silicates is the SiO4, tetrahedron., (d) Feldspars are not aluminosilicates., (Karnataka NEET 2013), 27. Name the two types of the structure of silicate in, which one oxygen atom of [SiO 4]4– is shared?, (a) Linear chain silicate (b) Sheet silicate, (c) Pyrosilicate, (d) Three dimensional, (2011), 28. The straight chain polymer is formed by, (a) hydrolysis of CH3SiCl3 followed by condensation, polymerisation, (b) hydrolysis of, (CH3)4Si, by addition, polymerisation, (c) hydrolysis of (CH3)2SiCl2 followed by, condensation polymerisation, (d) hydrolysis of (CH3)3SiCl followed by, condensation polymerisation., (2009), , www.mediit.in

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The p-Block Elements, , 85, , 29. Which of the following anions is present in the, chain structure of silicates?, (a) (Si O2–), (b) (SiO2–), 2, , 5, , (c) SiO4–, , n, , 3n, , (d) Si O6–, , 4, , to take up small molecules., (c) Zeolites are aluminosilicates having three, dimensional network., 4–, , (2007), , 27, , 30. Which one of the following statements about the, zeolite is false?, (a) They are used as cation exchangers., (b) They have open structure which enables them, , 5–, , (d) Some of the SiO4 units are replaced by AlO4, and AlO9– ions in zeolites., (2004), 6, , 31. The substance used as a smoke screen in warfare is, (a) SiCl4, (b) PH3, (c) PCl5, (d) acetylene., (1989), , ANSWER KEY, , 1., 11., 21., 31., , (d), (c), (c), (a), , 2., 12., 22., , (b), (b), (d), , 3., 13., 23., , (a), (a), (c), , 4., 14., 24., , (b), (d), (a), , 5., 15., 25., , (c), (a), (d), , 6., 16., 26., , (b), (a), (d), , 7., 17., 27., , (b), (a), (c), , 8., 18., 28., , (c), (c), (c), , 9., 19., 29., , (c), (c), (b), , 10., 20., 30., , (a), (b), (d), , Hints & Explanations, 1. (d), 2. (b) : AlF3 is insoluble in anhydrous HF because, the F– ions are not available in hydrogen bonded HF, molecules but, it becomes soluble in presence of little, amount of KF due to formation of complex, K3[AlF6]., AlF3 + 3KF  K3[AlF6], 3. (a) : In group 13 elements, stability of +3 oxidation, state decreases down the group while that of +1 oxidation, state increases due to inert pair effect. Hence, stability of, +1 oxidation state increases in the sequence :, Al < Ga < In < Tl., 4. (b) : AlCl3 forms a dimer, as Al due to the presence, of 3d-orbitals can expand its covalency from four to six., Also dimerisation enables Al atoms to complete their, octets., , 5. (c) : Boron does not have vacant d-orbitals in its, valence shell, so it cannot expand its covalency beyond 4, i.e., ‘B’ cannot form the ions like MF63–., 6. (b) : The relative Lewis acid character of boron, trihalides is found to follow the following order,, BBr3 > BCl3 > BF3, but the expected order on the basis, of electronegativity of the halogens (electronegativity of, halogens decreases from F to I) should be, BF 3 > BCl3 >, BBr3., This anomaly is explained on the basis of the relative, , www.neetujee.com, , tendency of the halogen atom to back donate its, unutilised electrons to vacant p-orbital of boron atom., In BF3, boron has a vacant 2p-orbital and each fluorine, has fully filled unutilised 2p-orbitals. Fluorine transfers, two electrons to vacant 2p-orbital of boron, thus forming, p-p bond., , This type of bond has some double bond character and, is known as dative or back bonding. All the three bond, lengths are same. It is possible when double bond is, delocalized. The delocalization may be represented as :, , The tendency to back donate decreases from F to I as, energy level difference between B and halogen atom, increases from F to I. So, the order of Lewis acid strength, is BF3 < BCl3 < BBr3., 7. (b) : Lewis acids are those substances which can, accept a pair of electrons and boron compounds usually, are deficient in electrons., 8. (c) : Boric acid behaves as a Lewis acid, by accepting, a pair of electrons from OH– ion of water thereby, releasing a proton., , www.mediit.in

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86, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 9. (c) : BN is known as inorganic graphite and has, structure similar to graphite., , 21. (c) : All the above are conductors except diamond., Diamond is an insulator., , 10. (a) : Each ‘B’ atom in diborane (B2H6) is sp3hybridised. Of the 4-hybrid orbitals, three have one, electron each, while the 4th is empty. Two orbitals of, each form  bonds with two ‘H’-atoms, while one of, the remaining hybrid orbital (either filled or empty), 1s, orbital of ‘H’ atom and one of the hybrid orbitals of other, ‘B’ atom overlap to form three centered two electron, bond. So there exists two such type of three centered, bonds., 11. (c) : H3BO3 is a weak monobasic acid., B(OH)3 + H2O  [B(OH)4]– + H+, , 22. (d) : (A) Solid CO2 (dry ice) is used as refrigerant, for ice-cream and frozen food., (B) The structure of C60 contains twenty six-membered, rings and twelve five-membered rings., (C) and (D) are correct statements., , 12. (b) : Generally halides of group-14 elements are, covalent in nature. PbF4 and SnF4 are exceptions which, are ionic in nature., 13. (a) : [SiCl6]2– is not stable due to steric hindrance by, large sized Cl atoms., 14. (d) : The inertness of s-subshell electrons towards, bond formation is known as inert pair effect. This effect, increases down the group thus, for Sn, +4 oxidation state, is more stable, whereas, for Pb, +2 oxidation state is more, stable,, Sn2+ is, while, Pb4+ is oxidising., 2, 15. (a)i.e.,, : When, nsreducing, electrons, of outermost, shell do not, participate in bonding then these ns2 electrons are called, inert pair and the effect is called inert pair effect. Due, to this inert pair effect Ge, Sn and Pb of group 14 have, a tendency to form both +4 and +2 ions. Now the inert, pair effect increases down the group, hence the stability, of M2+ ions increases and M 4+ ions decreases down the, group. For this reason, Pb2+ is more stable than Pb4+ and, Sn4+ is more stable than Sn2+., 16. (a) : Carbon has no d-orbitals, while silicon, contains d-orbitals in its valence shell which can be used, for bonding purposes., 17. (a) : Except diamond other three conduct electricity., Potassium and sodium are metallic conductors, while, graphite is a non-metallic conductor., 18. (c) : Lead pencil contains graphite and clay. It does, not contain lead., 19. (c) : In graphite each carbon atom undergoes, sp2-hybridisation and is covalently bonded to three other, carbon atoms by single bonds. The fourth electron forms, -bond. The electrons are delocalised over the whole, sheet i.e. electrons are spread out between the structure., 20. (b) : In diamond, each carbon atom is sp3 hybridized, and thus, forms covalent bonds with four other carbon, atoms lying at the corners of a regular tetrahedron., , 23. (c) : Silicones being biocompatible are used in, surgical and cosmetic plants., 24. (a) : It can form only dimer., 25. (d) : SiO44– orthosilicate is basic unit of silicates., 26. (d) : Feldspars are three dimensional aluminosilicates., 27. (c) : Pyrosilicate contains two units of SiO4– joined, 4, , along a corner containing oxygen atom., , 28. (c) : Hydrolysis of substituted chlorosilanes yields, corresponding silanols which undergo polymerisation., Out of the given chlorosilanes, only (CH3)2SiCl2 will give, linear polymer on hydrolysis followed by polymerisation., , 29. (b) : Chain silicates are formed by sharing two, oxygen atoms by each tetrahedra. Anions of chain silicate, have two general formula :, 6n–, (i) (SiO3)n2n– (ii) (Si 4O 11 )n, 30. (d) : In zeolites, some of the Si 4+ ions may be, replaced by Al3+ ions. This results in unbalanced anionic, charge. To maintain electrical neutrality, positive ions, must be introduced., , 31. (a) : SiCl4 gets hydrolysed in moist air and, gives white fumes which are used as a smoke screen in, warfare., , , , www.neetujee.com, , www.mediit.in

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Organic Chemistry - Some Basic Principles and Techniques, , 89, , 27. The IUPAC name of, , 21. Name of the compound given below is, , is, (a), (b), (c), (d), (a), (b), (c), (d), , 4-ethyl-3-methyloctane, 3-methyl-4-ethyloctane, 2,3-diethylheptane, 5-ethyl-6-methyloctane., , (1992), , 12.6 Isomerism, (2003), , 22. IUPAC name of the following is, CH2 CH – CH2 – CH2 – C CH, (a) 1,5-hexenyne, (b) 1-hexene-5-yne, (c) 1-hexyne-5-ene, (d) 1,5-hexynene., (2002), 23. The incorrect IUPAC name is, (a), - 2-methyl-3-butanone, (b), , 4-hydroxy-1-methylpentanal, 4-hydroxy-2-methylpent-2-en-1-al, 2-hydroxy-4-methylpent-3-en-5-al, 2-hydroxy-3-methylpent-2-en-5-al., , -2,3-dimethylpentane, , 28. Which among the given molecules can exhibit, tautomerism?, , (a) III only, (c) Both I and II, , (b) Both I and III, (d) Both II and III, (NEET-II 2016), , 29. Given :, , (c) CH3 – C CCH(CH3)2 -4-methyl-2-pentyne, (d), , -3-bromo-2-chlorobutane., (2001), , 24. The IUPAC name of (CH3)2CH – CH2 – CH2Br is, (a) 1-bromo-3-methylbutane, (b) 2-methyl-3-bromopropane, (c) 1-bromopentane, (d) 2-methyl-4-bromobutane., (1996), 25. The IUPAC name for, CH3CH CHCH2CHCH2COOH is, |, NH2, (a) 3-amino-5-heptenoic acid, (b) -amino--heptenoic acid, (c) 5-amino-2-heptenoic acid, (d) 5-amino-hex-2-enecarboxylic acid., , (a), (b), (c), (d), (1995), , 26. 2-Methyl-2-butene will be represented as, (a), (b), (c), (d), , www.neetujee.com, , Which of the given compounds can exhibit, tautomerism?, (a) II and III, (b) I, II and III, (c) I and II, (d) I and III, (2015, Cancelled), 30. The enolic form of ethyl acetoacetate as shown, below has, , (1992), , 9 sigma bonds and 2 pi-bonds, 9 sigma bonds and 1 pi-bond, 18 sigma bonds and 2 pi-bonds, 16 sigma bonds and 1 pi-bond. (2015, Cancelled), , 31. Which one of the following pairs represents, stereoisomerism?, (a) Structural isomerism and geometrical isomerism, (b) Optical isomerism and geometrical isomerism, (c) Chain isomerism and rotational isomerism, (d) Linkage isomerism and geometrical isomerism, (2005), 32. The molecular formula of diphenylmethane,, is C13H12., How many structural isomers are possible when one, of the hydrogen is replaced by a chlorine atom?, (a) 6, (b) 4, (c) 8, (d) 7 (2004), , www.mediit.in

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90, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 33. Tautomerism is exhibited by, (a) R3CNO2, (b) RCH2NO2, (c) (CH3)3CNO, (d) (CH3)2NH, (1997), 34. The number of isomers in C4H10O will be, (a) 7, (b) 8, (c) 5, (d) 6 (1996), , can form a bond by accepting a pair of electrons, from a nucleophile, (c) electrophile can be either neutral or positively, charged species and can form a bond by, accepting a pair of electrons from a nucleophile, (d) electrophile is a negatively charged species and, can form a bond by accepting a pair of electrons, from a nucleophile., (NEET 2017), , 35. Isomers of a substance must have the same, (a) structural formula (b) physical properties, (c) chemical properties (d) molecular formula., (1991), 36. How many chain isomers could be obtained from, the alkane C6H14?, (a) Four, (b) Five, (c) Six, (d) Seven, (1988), , 12.7 Fundamental Concepts in Organic, Reaction Mechanism, 37. A tertiary butyl carbocation is more stable than a, secondary butyl carbocation because of which of the, following?, (a) –I effect of – CH3 groups, (b) +R effect of – CH3 groups, (c) –R effect of – CH3 groups, (d) Hyperconjugation, (NEET 2020), , 42. Which of the following statements is not correct for, a nucleophile?, (a) Ammonia is a nucleophile., (b) Nucleophiles attack low e– density sites., (c) Nucleophiles are not electron seeking., (d) Nucleophile is a Lewis acid., (2015), 43. Treatment of cyclopentanone, , with methyl, , lithium gives which of the following species?, (a) Cyclopentanonyl radical, (b) Cyclopentanonyl biradical, (c) Cyclopentanonyl anion, (d) Cyclopentanonyl cation, (2015, Cancelled), 44. Which of the following is the most correct electron, displacement for a nucleophilic reaction to take place?, (a), , 38. The most stable+carbocation, among the following is, (a) (CH ) C—CH—CH, 3 3, , 3, , (b), , (b) CH 3—CH2—C+ H—CH2—CH 3, (c) CH 3—C+ H—CH 2—CH 2—CH 3, (d) CH —CH —C+ H, (Odisha NEET 2019), 3, , 2, , (c), , 2, , 39. Which of the following is correct with respect to – I, effect of the substituents? (R = alkyl), (a) – NH2 < – OR < – F (b) – NR2 < – OR < – F, (c) – NH2 > – OR > – F (d) – NR2 > – OR > – F, (NEET 2018, 1998), , (2015, Cancelled), , (d), , 45. Consider the following compounds :, , 40. Which of the following carbocations is expected to, be most stable?, (a), , (b), , (c), , (d), , Hyperconjugation occurs in, (a) III only, (b) I and III, (c) I only, (d) II only., (2015, Cancelled), 46. In which of the following compounds, the C—Cl, bond ionisation shall give most stable carbonium, ion?, (NEET 2018), , (a), , (b), , 41. The correct statement regarding electrophile is, (a) electrophile is a negatively charged species and, can form a bond by accepting a pair of electrons, from another electrophile, (b) electrophiles are generally neutral species and, , (c), , (d), , www.neetujee.com, , (2015, Cancelled), , www.mediit.in

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92, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 60. The most suitable method of separation of 1 : 1, mixture of ortho and para-nitrophenols is, (a) chromatography, (b) crystallisation, (c) steam distillation, (d) sublimation., (NEET 2017, 1999, 1993), , (a) sodamide, (c) sodium nitrite, , 61. The best method for the separation of naphthalene, and benzoic acid from their mixture is, (a) distillation, (b) sublimation, (c) chromatography (d) crystallisation. (2005), 62. In steam distillation of toluene, the pressure of, toluene in vapour is, (a) equal to pressure of barometer, (b) less than pressure of barometer, (c) equal to vapour pressure of toluene in simple, distillation, (d) more than vapour pressure of toluene in simple, distillation., (2001), 63. Which of the following technique is most, suitable for purification of cyclohexanone from a, mixture containing benzoic acid, isoamyl alcohol,, cyclohexane and cyclohexanone?, (a) Sublimation, (b) Evaporation, (c) Crystallisation (d) IR spectroscopy, (1997), , 12.9 Qualitative Analysis of Organic, Compounds, 64. Nitrogen detection in an organic compound is, carried out by Lassaigne’s test. The blue colour, formed corresponds to which of the following formulae?, (a) Fe3[Fe(CN)6]2, (b) Fe4[Fe(CN)6]3, (c) Fe4[Fe(CN)6]2, (d) Fe3[Fe(CN)6]3, (Karnataka NEET 2013), 65. The Lassaigne’s extract is boiled with conc. HNO3, while testing for halogens. By doing so it, (a) decomposes Na2S and NaCN, formed, (b) helps in the precipitation of AgCl, (c) increases the solubility product of AgCl, (d) increases the concentration of NO3– ions., (2011), 66. In sodium fusion test of organic compounds, the, nitrogen of the organic compound is converted into, , (b) sodium cyanide, (d) sodium nitrate., (1991), , 67. Lassaigne’s test is used in qualitative analysis to, detect, (a) nitrogen, (b) sulphur, (c) chlorine, (d) all of these., (1989), 68. A blue colouration is not obtained when, (a) ammonium hydroxide dissolves in copper, sulphate, (b) copper sulphate solution reacts with, K4[Fe(CN)6], (c) ferric chloride reacts with sod. ferrocyanide, (d) anhydrous CuSO4 is dissolved in water. (1989), , 12.10 Quantitative Analysis, 69. In Duma’s method for estimation of nitrogen, 0.25 g, of an organic compound gave 40 mL of nitrogen, collected at 300 K temperature and 725 mm, pressure. If the aqueous tension at 300 K is 25 mm,, the percentage of nitrogen in the compound is, (a) 16.76 (b) 15.76 (c) 17.36 (d) 18.20, (2015, Cancelled), 70. In the Kjeldahl’s method for estimation, of, nitrogen present in a soil sample, ammonia, evolved from 0.75 g of sample neutralized, 10 mL of 1 M H2SO4. The percentage of nitrogen in, the soil is, (a) 37.33, (b) 45.33, (c) 35.33, (d) 43.33, (2014), 71. In Dumas’ method of estimation of nitrogen 0.35 g, of an organic compound gave 55 mL of nitrogen, collected at 300 K temperature and 715 mm, pressure. The percentage composition of nitrogen in, the compound would be (aqueous tension at 300 K, = 15 mm), (a) 15.45, (b) 16.45, (c) 17.45, (d) 14.45, (2011), 72. Kjeldahl’s method is used in the estimation of, (a) nitrogen, (b) halogens, (c) sulphur, (d) oxygen., (1990), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., 61., 71., , (b), (d), (a), (b), (c), (c), (b), (b), , www.neetujee.com, , 2., 12., 22., 32., 42., 52., 62., 72., , (b), (c), (b), (b), (d), (b), (b), (a), , 3., 13., 23., 33., 43., 53., 63., , (d), (d), (a), (b), (c), (c), (d), , 4., 14., 24., 34., 44., 54., 64., , (d), (a), (a), (a), (a), (c), (b), , 5., 15., 25., 35., 45., 55., 65., , (d), (a), (a), (d), (a), (a), (a), , 6., 16., 26., 36., 46., 56., 66., , (d), (a), (b), (b), (d), (d), (b), , 7., 17., 27., 37., 47., 57., 67., , (d), (b), (b), (d), (c), (c), (d), , 8., 18., 28., 38., 48., 58., 68., , (c), (c), (a), (c), (a), (d), (b), , 9., 19., 29., 39., 49., 59., 69., , (a), (c), (b), (a,b), (a), (b), (a), , 10., 20., 30., 40., 50., 60., 70., , (a), (a), (c), (c), (b), (c), (a), , www.mediit.in

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93, , Organic Chemistry - Some Basic Principles and Techniques, , Hints & Explanations, 5, , 1., , 4, , 3, , 2, , 1, , (b) : HC C — CH CH — CH3, , 14. (a) :, , Pent-2-en-4-yne, (10 -bonds and 3-bonds), , 2., , (b) :, , 3. (d) : There are four double bonds. Hence, no. of, -electrons = 2 × 4 = 8., 4., , CH3 2 1, 3, CH – CH – C CH, , CH3, 7, 6, 5, (d) : CH – C – C, H, sp3, , 3, , 15. (a) :, 4-Ethyl-3-propylhex-1-ene, , 4, , sp3 sp2, , sp2, , sp3, , sp, , sp, , CH3, , , C2 - sp, C3 - sp3, C5 - sp2 and C6 - sp3, sp3, , 5., , sp2, , (d) : CH3 CH, 6, , 5, , sp2, , sp3, , sp, , 3, , 2, , 17. (b) :, 1, , The state of hybridisation of carbon in 1, 3 and 5 position, are sp, sp3 and sp2., 6., , (d) :, , 7., , (d) :, , 8., , (c) :, , C, , sp, , CH CH2 C CH, 4, , 16. (a) :, , If a molecule contains both carbon-carbon double or, triple bonds, the two are treated as per in seeking the, lowest number combination. However, if the sum of, numbers turns out to be the same starting from either, of the carbon chain, then lowest number is given to the, C C double bond., 18. (c) :, , 9. (a) : Tetrachloroethene being an alkene has sp2hybridised C-atoms and hence the angle Cl – C – Cl is, 120° while in tetrachloromethane, carbon is sp3, hybridised, therefore the angle Cl – C – Cl is 109°28, , Since the sum of numbers starting from either side of the, carbon chain turns out to be the same, so lowest number, is given to the C C end., 19. (c) :, , 10. (a) :, , It is 2,3-dimethylpentanoyl chloride., , (C6H7O2N), Hence, it is homocyclic (as the ring system is made of, one type of atoms, i.e., carbon) but not aromatic., , 20. (a) :, 4-Ethyl-3-methylheptane, , 11. (d) :, 21. (a) :, —, , CH3, 12. (c) : CH3 — CH — CH2 —, , E, , (iso-butyl group), , 13. (d) :, IUPAC name of the compound is, 3- ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid., , www.neetujee.com, , 22. (b) :, The double bond gets priority over triple bond. Therefore,, correct IUPAC name is 1-hexene-5-yne., 23. (a) :, , www.mediit.in

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Organic Chemistry - Some Basic Principles and Techniques, , 95, , 52. (b) : +R effect of —OH group is greater than that, of —OCH3 group., , carbocation. As is consists of maximum number of, -hydrogens and stablised by hyperconjugation., 39. (a, b) : –I effect increases on increasing the, electronegativity of atom.,  – NH2 < – OR < – F (–I effect), Also, – NR2 < – OR < – F (–I effect), 40. (c) : –NO2 group is meta-directing, thus will, stabilize a electrophile at m-position., 41. (c), 42. (d) : Nucleophiles are electron rich species hence,, they are Lewis bases., , 53. (c), 54. (c) : Higher the no. of electron releasing groups, lower will be stability of carbanion, and vice-versa. So,, the order of stability of carbanions is, , 55. (a) : In case of different nucleophiles, but present in, the same group in the periodic table, then larger is the, atomic mass, higher is the nucleophilicity. Hence, the, increasing order of nucleophilicity of the halide ions is, F– < Cl– < Br– < I–., , 43. (c) :, , 56. (d) :, , 3° carbon is more stable due to the stabilization of the, charge by three methyl groups (or +I – effect). It can also, be explained on the basis of hyperconjugation. Greater, the number of hyperconjugative -H atoms, more will, be the hyperconjugative structures and more will be the, stability., , 44. (a) : Nucleophile will attack a stable carbocation, (SN1 reaction)., , 45. (a) : Hyperconjugation can occur only in compound, III as it has -hydrogen atom., 46. (d) :, , is, , most, , stable, , due, , to, , hyperconjugation., 47. (c), 48. (a) : Greater the number of electron donating alkyl, groups (+I effect), greater is the stability of carbocations., +I effect is in the order :, , 57. (c) : 3º > 2º > 1º more the delocalisation of positive, charge, more is its stability., 58. (d) : All the properties mentioned in the question, suggest that it is a benzene molecule. Since in benzene all, carbons are sp2-hybridised, therefore, C – C – C angle is, 120º., 59. (b) : Paper chromatography is a type of partition, chromatography., 60. (c) : The o- and p-nitrophenols are separated by, steam distillation since o-isomer is steam volatile due to, intramolecular H-bonding while p-isomer is not steam, volatile due to association of molecules by intermolecular, H-bonding., , More the number of hyperconjugative structures of, carbocations, more is the stability., Hence, the order of stability of carbocations is, 5 < 4 < 3 < 1 < 2., 49. (a), 50. (b) : More the number of hyperconjugative, structures, the greater is the stability., 51. (c) : Nucleophilic substitution reaction involves the, displacement of a nucleophile by another., , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 13, , Hydrocarbons, (d) the eclipsed conformation of ethane is more, stable than staggered conformation, because, eclipsed conformation has no torsional strain., (NEET-I 2016), , 13.2 Alkanes, 1., , 2., , 3., , 4., , 5., , Which of the following alkane cannot be made in, good yield by Wurtz reaction?, (a) n-Hexane, (b) 2, 3-Dimethylbutane, (c) n-Heptane, (d) n-Butane, (NEET 2020), , 6., , The alkane that gives only one monochloro product, on chlorination with Cl 2 in presence of diffused, sunlight is, (a) 2,2-dimethylbutane (b) neopentane, (c) n-pentane, (d) isopentane., (Odisha NEET 2019), Hydrocarbon (A) reacts with bromine by, substitution to form an alkyl bromide which by, Wurtz reaction is converted to gaseous hydrocarbon, containing less than four carbon atoms. (A) is, (a) CH CH, (b) CH2 CH2, (c) CH3 — CH3, (d) CH4, (NEET 2018), With respect to the conformers of ethane, which of, the following statements is true?, (a) Bond angle changes but bond length remains, same., (b) Both bond angle and bond length change., (c) Both bond angle and bond length remain same., (d) Bond angle remains same but bond length, changes., (NEET 2017), The correct statement regarding the comparison of, staggered and eclipsed conformations of ethane, is, (a) the eclipsed conformation of ethane is more, stable than staggered conformation even though, the eclipsed conformation has torsional strain, (b) the staggered conformation of ethane is more, stable than eclipsed conformation, because, staggered conformation has no torsional strain, (c) the staggered conformation of ethane is less, stable than eclipsed conformation, because, staggered conformation has torsional strain, , www.neetujee.com, , In the following the most stable conformation of, n-butane is, CH3, CH3, H, H, H, CH3, (a), (b), H, H, H, H, CH3, H, H3 C, , H3C, , CH 3, (d) H, H, , (c) H, H, , H, H, , H, (2010), , H, H3C, , 7., , Liquid hydrocarbons can be converted to a mixture, of gaseous hydrocarbons by, (a) oxidation, (b) cracking, (c) distillation under reduced pressure, (d) hydrolysis., (2010), , 8., , Which of the following conformers for ethylene, glycol is most stable?, OH, , OH, OH, , (a), , (b), H, , H, H, , H, OH, OH, H, , (c), , HH, OH, OH, , (d), HO, H, , HH, , H, , H, H, , (Mains 2010), , www.mediit.in

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98, , 9., , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , Dihedral angle in staggered form of ethane is, (a) 0°, (b) 120°, (c) 60°, (d) 180°, , (b), (2000), , 10. Which of the following reaction is expected to, readily give a hydrocarbon product in good yields?, C H OH, (b) (CH ) CHCl 2 5, (a) CH CH Cl2, 3, , h3 2, Electrolysis, Oxidation, , 3, , (c) RCOOK, , (d) RCOOAg, , I2, , (1997), , 11. In the commercial gasolines, the type of, hydrocarbons which are more desirable is, (a) linear unsaturated hydrocarbon, (b) toluene, (c) branched hydrocarbon, (d) straight-chain hydrocarbon., (1997), 12. The most stable conformation of n-butane is, (a) gauche, (b) staggered, (c) skew boat, (d) eclipsed., (1997), 13. Which of the following is used as an antiknocking, material?, (a) Glyoxal, (b) Freon, (c) T.E.L., (d) Ethyl alcohol (1996), 14. Reactivity of, hydrogen, atoms, attached, to different carbon atoms in alkanes has the order, (a) tertiary > primary > secondary, (b) primary > secondary > tertiary, (c) both (a) and (b), (1993), (d) tertiary > secondary > primary., , 13.3 Alkenes, 15. An alkene on ozonolysis gives methanal as one of, the product. Its structure is, CH2 CH2 CH3, CH CH CH3, (a), , (b), CH2, , (c), , CH, , CH2, , (c), , CH2CH2CH3, , (NEET 2019), , (d), , 17. The most suitable reagent for the following, conversion, is, , (a) Hg2+/H+, H2O, (b) Na/liquid NH3, (c) H2, Pd/C, quinoline (d) Zn/HCl (NEET 2019), 18. Which of the following compounds shall not, produce propene by reaction with HBr followed by, elimination or direct only elimination reaction?, H2 C, CH2, (a), C, H2, H2, (b) H3 C C CH2OH, (c) H2C, C O, H2, (d) H3 C C CH2Br, (NEET-II 2016), 19. The compound that will react most readily with, gaseous bromine has the formula, (a) C3 H6, (b) C2 H2, (c) C4 H10, (d) C2 H4, (NEET-II 2016), , 20. In the reaction with HCl, an alkene reacts in, accordance with the Markovnikov’s rule to give, a product 1-chloro-1-methyl-cyclohexane. The, possible alkene is, CH2, CH3, (a), , (A), , (b), , (d), , (B), CH3, , (NEET 2020), 16. An alkene A on reaction with O3 and Zn H2O gives, propanone and ethanal in equimolar ratio. Addition, of HCl to alkene A gives B as the major product. The, structure of product B is, (a), , www.neetujee.com, , (c) (A) and (B), , (d), , (2015), , 21. Which of the following is not the product of, dehydration of, (a), , OH, , ?, , (b), , www.mediit.in

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Hydrocarbons, , 101, , 49. Which of the following reagents will be able to, distinguish between 1-butyne and 2-butyne?, (a) NaNH2, (b) HCl, (c) O2, (d) Br2, (Mains 2012), , 55. When acetylene is passed through dil. H2SO4 in the, presence of HgSO4, the compound formed is, (a) acetic acid, (b) ketone, (c) ether, (d) acetaldehyde. (1999), , 50. Considering the state of hybridization of carbon, atoms, find out the molecule among the following, which is linear?, (a) CH3 — CH CH — CH3, (b) CH3 — C C — CH3, (c) CH2 CH — CH2 — C CH, (d) CH3 — CH2 — CH2 — CH3, (2011), , 56. The cylindrical shape of an alkyne is due to, (a) two sigma C – C and one  C – C bonds, (b) one sigma C – C and two  C – C bonds, (c) three sigma C – C bonds, (d) three  C – C bonds., (1997), , 51. Base strength of, is in the order of, (a) (i) > (iii) > (ii), (c) (ii) > (i) > (iii), , (b) (i) > (ii) > (iii), (d) (iii) > (ii) > (i) (2008), , 52. Predict the product C obtained in the following, reaction of 1-butyne., HI, , CH, CH, 3, 2, C, , C, , B, , CH+ HCl, , Reagent, , 57. R – CH2 – CCl2 – R, The reagent is, (a) Na, (c) KOH in C2H5OH, , R–CC–R, , (b) HCl in H2O, (d) Zn in alcohol. (1993), , 58. A compound is treated with NaNH2 to give sodium, salt. Identify the compound., (a) C2H2, (b) C6H6, (c) C2H6, (d) C2H4, (1993), 59. The shortest C–C bond distance is found in, (a) diamond, (b) ethane, (c) benzene, (d) acetylene., (1991), 60. Acetylenic hydrogens are acidic because, , I, (a) CH3 CH2 CH2 C H, Cl, I, (b) CH3 CH2 CH CH2Cl, I, (c) CH3CH2 C CH3, Cl, (2007), , (d) CH3 CH CH2CH2 I, Cl, 53. Products of the following reaction :, (i) O3, CH C CCH CH, 2, , 3, , are, , 3, , (ii) hydrolysis, , (a) CH3COOH + CO2, (b) CH3COOH + HOOCCH2CH3, (c) CH3CHO + CH3CH2CHO, (d) CH3COOH + CH3COCH3, (2005), 54. When CH CH CHCl is treated with NaNH , the, 3, , 2, , 2, , product formed is, (a) CH3 – CH CH2, (b) CH3 – C, NH2, (c) CH CH CH, 3, , 2, , (d) CH3CH2CH, , www.neetujee.com, , 61. Which is the most suitable reagent among the, following to distinguish compound (3) from rest of, the compounds?, (1) CH3 – C, C – CH3, (2) CH3 – CH2 – CH2 – CH3, (3) CH3 – CH2C CH, (4) CH3 – CH, CH2, (a) Bromine in carbon tetrachloride, (b) Bromine in acetic acid, (c) Alk. KMnO4, (d) Ammoniacal silver nitrate, (1989), , 2, , 13.5 Aromatic hydrocarbons, CH, , 62. Among the following the reaction that proceeds, through an electrophilic substitution is, , NH2, Cl, NH2, , (a) sigma electron density of C – H bond in, acetylene is nearer to carbon, which has 50%, s-character, (b) acetylene has only open hydrogen in each, carbon, (c) acetylene contains least number of hydrogens, among the possible hydrocarbons having two, carbons, (d) acetylene belongs to the class of alkynes with, molecular formula, CnH2n–2., (1989), , (a), (2002), , www.mediit.in

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Hydrocarbons, (c), , 103, , (d), S, (Karnataka NEET 2013), , 73. Among the following compounds the one that is, most reactive towards electrophilic nitration is, (a) benzoic acid, (b) nitrobenzene, (c) toluene, (d) benzene., (2012, 1992), 74. The reaction of toluene with Cl2 in presence of FeCl3, gives X and reaction in presence of light gives Y., Thus, X and Y are, (a) X = benzal chloride,, Y = o-chlorotoluene, (b) X = m-chlorotoluene,, Y = p-chlorotoluene, (c) X = o- and p-chlorotoluene,, Y = trichloromethyl benzene, (d) X = benzyl chloride,, Y = m-chlorotoluene., (2010), 75. Benzene reacts with CH3Cl in the presence of, anhydrous AlCl3 to form, (a) chlorobenzene, (b) benzyl chloride, (c) xylene, (d) toluene., (2009), 76. Nitrobenzene can be prepared from benzene by, using a mixture of conc. HNO3 and conc. H2SO4. In, the mixture, nitric acid acts as a/an, (a) acid, (b) base, (c) catalyst, (d) reducing agent., (2009), 77. Which one of the following is most reactive towards, electrophilic attack?, (a), , (b), , (c), , (d), , (2004), , 80. Which one of the following is a free-radical, substitution reaction?, (a), , CH3, , + Cl2, , Boiling, , CH2Cl, , (b), (c), , CH2NO2, , CH2Cl + AgNO, 2, , (d) CH3CHO + HCN → CH3CH(OH)CN, , (2003), , 81. The correct order of reactivity towards the, electrophilic substitution of the compounds aniline, (I), benzene (II) and nitrobenzene (III) is, (a) III > II > I, (b) II > III > I, (c) I < II > III, (d) I > II > III, (2003), 82. Increasing order of electrophilic substitution for, following compounds, CH3, (I), , (II), OCH3, , CF3, , (III), , (IV), , (a) IV < I < II < III, (c) I < IV < III < II, , (b) III < II < I < IV, (d) II < III < I < IV, (2000), , 83. In Friedel-Crafts reaction, toluene can be prepared, by, (a) C6H6 + CH3Cl, (b) C6H5Cl + CH4, (c) C6H6 + CH2Cl2, (d) C6H6 + CH3COCl, (2000), (2008), , 78. The order of decreasing reactivity towards an, electrophilic reagent, for the following would be, (i) benzene, (ii) toluene, (iii) chlorobenzene, (iv) phenol, (a) (ii) > (iv) > (i) > (iii), (b) (iv) > (iii) > (ii) > (i), (c) (iv) > (ii) > (i) > (iii), (d) (i) > (ii) > (iii) > (iv), (2007), 79. Using anhydrous AlCl3 as catalyst, which one of the, following reactions produces ethylbenzene (PhEt)?, (a) H3C – CH2OH + C6H6, , www.neetujee.com, , (b) CH3 – CH CH2 + C6H6, (c) H2C CH2 + C6H6, (d) H3C – CH3 + C6H6, , 84. In Friedel-Crafts alkylation, besides AlCl3 the other, reactants are, (a) C6H6 + CH3Cl, (b) C6H6 + CH4, (c) C6H6 + NH3, (d) C6H6 + CH3COCl, (1999), 85. Which of the following compounds will be most, easily attacked by an electrophile?, (a), (c), , OH, , (b), (d), , Cl, CH3, (1999, 1998), , www.mediit.in

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104, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 86. Which one of these is not compatible with arenes?, (a) Electrophilic additions, (b) Delocalisation of -electrons, (c) Greater stability, (d) Resonance, , (1998), , 87. Among the following compounds (I-III) the correct, order of reaction with electrophile is, , 89. The reactive species in the nitration of benzene is, (a) NO3+, (b) HNO, 3, (c) NO, (d) NO–, (1994), 2, , 2, , 90. Which is the correct symbol relating the two Kekule, structures of benzene?, (a), (b), (c), (d), (1993), 91. Select the true statement about benzene amongst, , (a) I > II > III, (c) II > III > I, , (b) I = II > III, (d) III < I < II, , the following, (a) because of unsaturation benzene easily, undergoes addition, (b) there are two types of C – C bonds in benzene, molecule, (c) there is cyclic delocalisation of -electrons in, benzene, (d) monosubstitution of benzene gives three, isomeric products., (1992), , (1997), , 88. Electrophile in the case of chlorination of benzene, in the presence of FeCl3 is, (a) Cl, (b) FeCl3, +, (c) Cl, (d) Cl–, (1996), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., 61., 71., 81., 91., , (c), (c), (a), (b), (c), (b), (d), (a), (d), (c), , 2., 12., 22., 32., 42., 52., 62., 72., 82., , (b), (b), (a), (c), (a), (c), (c), (d), (a), , 3., 13., 23., 33., 43., 53., 63., 73., 83., , (d), (c), (d), (a), (a), (b), (d), (c), (a), , 4., 14., 24., 34., 44., 54., 64., 74., 84., , (c), (d), (d), (a), (c), (b), (a), (c), (a), , 5., 15., 25., 35., 45., 55., 65., 75., 85., , (b), (c), (d), (d), (b), (d), (d), (d), (a), , 6., 16., 26., 36., 46., 56., 66., 76., 86., , (b), (d), (a), (d), (c), (b), (c), (b), (a), , 7., 17., 27., 37., 47., 57., 67., 77., 87., , (b), (c), (d), (d), (b), (c), (d), (a), (a), , 8., 18., 28., 38., 48., 58., 68., 78., 88., , (d), (c), (b), (c), (c), (a), (a), (c), (c), , 9., 19., 29., 39., 49., 59., 69., 79., 89., , (c), (a), (c), (c), (a), (d), (c), (c), (c), , 10., 20., 30., 40., 50., 60., 70., 80., 90., , (c), (c), (b), (d), (b), (a), (b), (a), (d), , Hints & Explanations, 1. (c) : Wurtz reaction is used for the preparation of, higher alkanes containing even number of C-atoms., Thus this reaction cannot be used for the preparation of, n-heptane., 2. (b) : In chlorination of alkanes, hydrogen is replaced, by chlorine. In neo-pentane, only one type of hydrogen, is present, hence replacement of any hydrogen atom will, give the same product., , 3. (d) : CH4, (A), , Br2/hv, , CH3 Br, , Na/dry ether, Wurtz reaction, , CH 3, , CH3, , (Less than 4, caron atoms), , 4. (c) : Conformers of ethane have different dihedral, angles but same bond angle and bond lengths., 5., , (b) :, , Angle of rotation or angle, of torsion or dihedral angle, , neo-pentane, , www.neetujee.com, , www.mediit.in

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Hydrocarbons, , 105, , Magnitude of torsional strain depends upon the angle of, rotation about C—C bond. Staggered form has the least, torsional strain and the eclipsed form has the maximum, torsional strain. So, the staggered conformation of ethane, is more stable than the eclipsed conformation., 6. (b) : The anti-conformation is the most stable, conformation of n-butane as in this, the bulky methyl, groups are as far apart as possible thereby keeping steric, repulsion at a minimum., 7. (b) : The process of cracking converts higher, alkanes into smaller alkanes and alkenes. This process, can be used for production of natural gas., 8. (d) : The conformation (d) is most stable because of, intermolecular H-bonding., 9. (c) : The staggered form of ethane has the following, structure and the dihedral angle is 60°, which means ‘H’, atoms are at an angle of 60° to each other., H, H, H, , . 60°, H, H, 10. (c) : When an aqueous solution of sodium or, potassium salt of carboxylic acid is electrolysed,, hydrocarbon is evolved at anode., 2RCOOK, ,  2K , , , H, , H, , H, CH 3, staggered, , H3C, CH3, , + HCHO, , H H, eclipsed, , H, H, , Methanal, , 16. (d) :, CH3, CH3, , CH, , C, , CH3, , HCl, , CH3, , C, , H, , CH2, , CH3, , Cl, , (A), , (B), , CH3, , O3 Zn/, H2O, , CH3, , C, , CH3 + CH3, , C, , O, , H, , O, , Addition of HCl to an alkene (A) will take place according, to Markownikoff’s rule., H3C, , H2, Pd/C, quinoline, , 17. (c) : CH3 C CCH3, , CH3, , CC, , H, , H, , cis-2-Butene, , H2 C CH 2 HBr CH3 CH 2 CH2 Elimination, C, Br, H2, H CCH CH, CH3CH2CH2OH, , CH2 C O, , HBr, , 3, , 2, , Elimination, , H3C–CH, , HBr, , CH3CH2CH2Br, , H2 C C OH, Elimination, , CH3CH, , CH2, O, , H3C C Br, CH2, , CH3, , 19. (a) : Rate of free radical substitution with Br2(g), depends upon the stability of free radical. Propenyl free, radical is allylic free radical which is more stable., , H, , 20. (c) :, , H3 C, H, , CHO, , Br, , 11. (c) : The branching of chain increases the octane, number of a fuel. High octane number means better fuel., 12. (b) : Newman projections for n-butane are, H 3C, , CH2, , (i) O3, (ii) Zn/H2 O, , –, , Alkane, , H, , CH2, , Cathode, , At anode : 2RCOO – 2e  R – R + 2CO2, –, , CH, , 18. (c) :, , H, , lectrolysis, E,  2RCOO, Oxidation, Anode, , CH2, 15. (c) :, , H, skew or gauche, , The staggered conformation has minimum repulsion, between the hydrogen atoms attached tetrahedrally, to the two carbon atoms. Thus, it is the most stable, conformation., 13. (c) : Tetraethyl lead (C2H5)4Pb, is used as an, antiknocking agent in gasoline used for running, automobiles., , 21. (a) :, , 14. (d) : The reactivity of H-atom depends upon the, stability of free radicals, therefore reactivity of H-atom, follows the order : 3º > 2º > 1º., , www.neetujee.com, , www.mediit.in

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Hydrocarbons, , 109, , some double bond character due to resonance of lone, pair of electrons with -bond., , CH3, , (no substitution occurs here), Br2, , 2, , 64. (a) : Biphenyl is coplanar as all C-atoms are sp, hybridised., 65. (d) : Pyrrole has maximum electron density on, 2 and 5. It generally reacts with electrophiles at the C2 or C-5 due to the highest degree of stability of the, protonated intermediate., H, 3, 2, , N1, H, , N, H+, , Y, +H, N, H, , (I), , (II), , (III), , CH3, , 4- Bromo-1,3-dimethyl, , 6-Bromo-1,3-dimethyl, benzene, (minor), , benzene, (major), , 70. (b) : —NO2 is most deactivating due to – I and – M, HH, Y + N, H, , N, H, (IV), , H, Y, , (V), , More stable ion, , +, , CH3, , Br, CH3, , Attack at position 3 or 4 yields a carbocation that is a, hybrid of structures (I) and (II). Attack at position 2 or 5, yields a carbocation that is a hybrid not only of structures, (III) and (IV) (analogous to I and II) but also of structure, (V). The extra stabilization conferred by (V) makes this, ion the more stable one. Also, attack at position 2 or 5 is, faster because the developing positive charge is, accommodated by three atoms of the ring instead of only, two., +, H (from HF), Addition reaction, , CH3, , Br, , +H, , H, N+ Y, H, , 66. (c) :, , 1,3-Dimethylbenzene, (m-Xylene), , H, Y, , 4, 5, , FeBr 3, , CH3, , effect., 71. (a) : Nitrobenzene is strongly deactivated, hence, will not undergo Friedel-Crafts reaction., 72. (d) : The molecules which do not satisfy Huckel rule, or (4n + 2)-electron rule are said to be non-aromatic., The compound (d) has total 4e–. It does not follow, (4n + 2) rule. So, it is non-aromatic compound., 73. (c) : As the +I effect increases reactivity towards, electrophilic reactions increases and as –I or –M effect, increases, reactivity towards electrophilic reactions, decreases. Thus, the order is, , +, , H, Friedel–Cra s, alkylation, , 67. (d) : Mechanism of nitration is :, HNO3 + 2H2SO4  NO+2+ 2HSO–4+ H3O+, If a large amount of KHSO4 is added then conc. of HSO4–, ions increases and the reaction will be shifted in backward, direction hence, the rate of nitration will be slower., 68. (a) :, , +9 O, 2, , CHCOOH, , V2O5 , 773 K, 2 –2CO2 , –H2O, , CHCOOH, , O, CHC, , Maleic acid, , O, , CHC, , , –H2 O, , O, Maleic anhydride, , 69. (c) : — CH3 group is o,p-directing. Because of, crowding, no substitution occurs at the carbon atom, between the two — CH3 groups in m-xylene, even though, two — CH 3 groups activate that position., , www.neetujee.com, , 74. (c) : The reaction of Cl2, in presence of FeCl3, with, toluene yields a ring substitution product., CH3, , CH3, , CH3, , Cl, , Cl2/FeCl3, , Cl, (X), , In presence of sunlight, free radical reaction takes, place., CH3, Cl2 /h, , CH2Cl, , CHCl2, , Cl2 /h, , CCl3, , Cl2 /h, , , (Y), , 75. (d) : This is Friedel-Crafts alkylation., CH3, + CH3Cl, Benzene, , AlCl3, Toluene, , www.mediit.in

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110, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , Mechanism: CH 3Cl + AlCl3  AlCl4– + CH3+, Lewis acid, , CH3, , +, , CH3, , AlCl –, , H, , + CH3, , substitution. While — CH3 and — OCH3 are electrondonating group which favours electrophilic substitution, in the benzene ring at ‘ortho’ and ‘para’ positions. The, +I effect of — OCH3 is more than — CH3, therefore the, correct order for electrophilic substitution is, CH3, OCH3, CF3, , Electrophile, , 4, , Toluene, , + AlCl3 + HCl, , <, , 76. (b) :, , (IV), , 77. (a) : Groups like, — Cl and — NO2 shows –I effect., –I groups attached to the benzene ring decrease the, electron density and hence less prone to electrophilic, attack. — OH not only shows –I effect but also +M, effect which predominates the –I character and electron, density is increased in the benzene ring which facilitates, electrophilic attack., 78. (c) : Benzene having any activating group i.e.,, — OH, — R undergoes electrophilic substitution easily, as compared to benzene itself. Thus, toluene and phenol, undergo electrophilic substitution easily. Chlorine due to, –I-effect deactivates the ring. So, it is difficult to carry out, the electrophilic substitution in chlorobenzene. Hence,, the order is C6H5OH > C6H5CH3 > C6H6 > C6H5Cl., 79. (c) : C6H5H + H2C == CH2, C6H5 CH2 CH, CH3, , 80. (a) :, , CH2, , heat, , 3, , ., , +H, Benzyl free radical, , Cl 2, , Energy, sunlight, , CH2, , 2Cl, , ., , CH2Cl, , + Cl, NH2, , NO2, , 81. (d) :, I, , II, , III, , — NH2 group is electron donating hence increases, electron density on ring. Benzene is also electron rich due, to delocalisation of electrons. — NO2 group is electron, withdrawing hence, decreases electron density on ring., Thus, correct order for electrophilic substitution is, I > II > III., 82. (a) : Due to –I effect of F atom, —CF3 on benzene, ring, deactivates the ring and does not favour electrophilic, , <, (I), , <, (II), , (III), , 83. (a) : In Friedel-Crafts reaction toluene is obtained, by the action of CH3Cl on benzene in presence of AlCl3., CH3, + CH3Cl, , AlCl3, , + HCl, Toluene, , 84. (a) : In Friedel-Crafts reaction, an alkyl group is, introduced into the benzene ring in presence of a Lewis, acid (AlCl3) catalyst. The reaction is, CH3, AlCl3, + CH3Cl, + HCl, Benzene, , Toluene, , 85. (a) : — OH, — Cl and — CH groups in benzene, 3, , are ortho, para directing groups. But among these, — OH group is strongly activating while — CH3 is weakly, activating and — Cl is deactivating. Thus, phenol will be, most easily attacked by an electrophile., 86. (a) : Arenes undergo electrophilic substitution, reactions and are resistant to addition reactions, due to, delocalisation of -electrons. These are also stabilized by, resonance., 87. (a) : In structure III, withdrawal of electrons by, —NO2 causes decrease in reaction rate while in structure I,, there is electron releasing effect by — OCH3 group which, accelerates the reaction., The order of reactivity towards electrophile is :, I > II > III, 88. (c) : Cl2 + FeCl3  FeCl4– + Cl+, 89. (c) : Nitronium ion (NO+) is2 an electrophile that, actually attacks the benzene ring., 90. (d) : Benzene shows Kekule structures which are, resonating structures and these structures are separated, by a double headed arrow ()., 91. (c) : Due to resonance all the C – C bonds in, the benzene possess same nature and the resonating, structures are obtained because of the delocalisation of, -electrons., , , www.neetujee.com, , www.mediit.in

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, , , , , , , CHAPTER, , 14, , Environmental, Chemistry, (c) Photochemical smog is formed through, photochemical reaction involving solar energy., (d) Photochemical smog does not cause irritation, in eyes and throat., (2012), , 14.2 Atmospheric Pollution, 1., , Which of the following is not correct about carbon, monoxide?, (a) It forms carboxyhaemoglobin., (b) It reduces oxygen carrying ability of blood., (c) The carboxyhaemoglobin (haemoglobin bound, to CO) is less stable than oxyhaemoglobin., (d) It is produced due to incomplete combustion., (NEET 2020), , 2., , Among the following, the one that is not a, greenhouse gas is, (a) sulphur dioxide (b) nitrous oxide, (c) methane, (d) ozone. (NEET 2019), , 3., , Which oxide of nitrogen is not a common pollutant, introduced into the atmosphere both due to natural, and human activity?, (a) N2O5, (b) NO2, (c) N2O, (d) NO (NEET 2018), , 4., , Which of the following is a sink for CO?, (a) Microorganisms present in the soil, (b) Oceans, (c) Plants, (d) Haemoglobin, (NEET 2017), , 5., , Which one of the following is not a common, component of photochemical smog?, (a) Ozone, (b) Acrolein, (c) Peroxyacetyl nitrate, (d) Chlorofluorocarbons, (2014), , 6., , Which one of the following statements regarding, photochemical smog is not correct?, (a) Carbon monoxide does not play any role in, photochemical smog formation., (b) Photochemical smog is an oxidising agent in, character., , www.neetujee.com, , 7., , Which one of the following is responsible for, depletion of the ozone layer in the upper strata of, the atmosphere?, (a) Polyhalogens, (b) Ferrocene, (c) Fullerenes, (d) Freons, (2004), , 8., , About 20 km above the earth, there is an ozone, layer. Which one of the following statements about, ozone and ozone layer is true?, (a) It is beneficial to us as it stops U.V. radiation., (b) Conversion of O3 to O2 is an endothermic, reaction., (c) Ozone is a triatomic linear molecule., (d) It is harmful as it stops useful radiation. (1995), , 14.3 Water Pollution, 9., , Which one of the following statements is not true?, (a) Clean water would have a BOD value of 5 ppm., (b) Fluoride deficiency in drinking water is, harmful. Soluble fluoride is often used to bring, its concentration upto 1 ppm., (c) When the pH of rain water is higher than 6.5, it, is called acid rain., (d) Dissolved Oxygen (DO) in cold water can reach, a concentration upto 10 ppm., (Karnataka NEET 2013), , 10. Which one of the following statement is not true?, (a) pH of drinking water should be between, 5.5 – 9.5., (b) Concentration of DO below 6 ppm is good for, the growth of fish., (c) Clean water would have a BOD value of less, than 5 ppm., , www.mediit.in

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112, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (d) Oxides of sulphur, nitrogen and carbon, are the, (2011), most widespread air pollutant., , (a), (b), (c), (d), , 14.7 Green Chemistry, 11. Green chemistry means such reactions which, , are related to the depletion of ozone layer, study the reactions in plants, produce colour during reactions, reduce the use and production of hazardous, chemicals., (2008), , ANSWER KEY, , 1., 11., , (c), (d), , 2., , (a), , 3., , (a), , 4., , (a), , 5., , (d), , 6., , (d), , 7., , (d), , 8., , (a), , 9., , (c), , 10., , (b), , Hints & Explanations, 1. (c) : The carboxyhaemoglobin is about 300 times, more stable than oxyhaemoglobin., 2. (a) : Besides carbon dioxide, ether greenhouse, gases are methane, water vapours, nitrous oxide, CFCs, and ozone., 3. (a), 4. (a) : Microorganisms present in the soil consume, atmospheric CO., 5., , (d), , radiations and break down liberating free atomic chlorine, which causes decomposition of ozone. This results in the, depletion of the ozone layer., , 8. (a) : Ozone layer is very beneficial to us, because it, stops harmful ultraviolet radiations to reach the earth., 9. (c) : When pH of rain water drops below 5.6 it is, called acid rain., , 6. (d) : Photochemical smog causes irritation in eyes, and throat., , 10. (b) : Fish dies in water bodes polluted by sewage, due to decrease in dissolved oxygen (D.O.), , 7. (d) : Chlorofluorocarbons such as freon-11 (CFCl3), and freon-12 (CF2Cl2) emitted as propellants in aerosol, spray cans, refrigerators, fire fighting reagents etc. are, stable compounds and chemically inert. They do not, react with any substance with which they come in contact, and thus float through the atmosphere unchanged and, eventually enter the stratosphere. There they absorb UV, , 11. (d) : Green chemistry is the design, development,, and implementation of chemical products and processes, to reduce or eliminate the use and generation of substances, hazardous to human health and the environment. Green, chemistry also refers to the redesign of chemical products, and processes with the goal of reducing or eliminating, any negative environmental or health effects., , , , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 1, , The Solid State, , 1.2 Amorphous and Crystalline Solids, 1., , 2., , 1.4 Crystal Lattice and Unit Cells, , The pure crystalline substance on being heated, gradually first forms a turbid liquid at constant, temperature and still at higher temperature, turbidity completely disappears. The behaviour is a, characteristic of substance forming, (a) allotropic crystals (b) liquid crystals, (c) isomeric crystals (d) isomorphous crystals., (1993), Glass is a, (a) liquid, (b) solid, (c) supercooled liquid, (d) transparent organic polymer., , For orthorhombic system axial ratios are, a ≠ b ≠ c and the axial angles are, (a)  =  =  ≠ 90º, (b)  =  =  = 90º, (c)  =  = 90º,   90º (d)    ≠   90º (1991), , 1.5 Number of Atoms in a Unit Cell, 8., , The number of carbon atoms per unit cell of, diamond unit cell is, (a) 6, (b) 1, (c) 4, (d) 8, (NEET 2013), , 9., (1991), , 3., , Most crystals show good cleavage because their, atoms, ions or molecules are, (a) weakly bonded together, (b) strongly bonded together, (c) spherically symmetrical, (d) arranged in planes., (1991), , 4., , The ability of a substance to assume two or more, crystalline structures is called, (a) isomerism, (b) polymorphism, (c) isomorphism, (d) amorphism., (1990), , 1.3 Classification of Crystalline Solids, 5., , Cation and anion combines in a crystal to form, following type of compound, (a) ionic, (b) metallic, (c) covalent, (d) dipole-dipole. (2000), , 6., , For two ionic solids CaO and KI, identify the wrong, statement among the following., (a) CaO has high melting point., (b) Lattice energy of CaO is much larger than that, of KI., (c) KI has high melting point., (d) KI is soluble in benzene., (1997), , www.neetujee.com, , 7., , In a face-centred cubic lattice, a unit cell is shared, equally by how many unit cells?, (a) 2, (b) 4, (c) 6, (d) 8, (2005), 10. When Zn converts from melted state to its solid, state, it has hcp structure, then find the number of, nearest atoms., (a) 6, (b) 8, (c) 12, (d) 4, (2001), 11. The fcc crystal contains how many atoms in each, unit cell?, (a) 6, (b) 8, (c) 4, (d) 5, (1996), 12. The number of atoms contained in a fcc unit cell of a, monatomic substance is, (a) 1, (b) 2, (c) 4, (d) 6, (1993), , 1.6 Closed Packed Structures, 13. A compound is formed by cation C and anion A., The anions form hexagonal close packed (hcp), lattice and the cations occupy 75% of octahedral, voids. The formula of the compound is, (a) C4A3, (b) C2A3, (c) C3A2, (d) C3A4 (NEET 2019), , www.mediit.in

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2, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 14. In calcium fluoride, having the fluorite structure,, the coordination numbers for calcium ion (Ca2+), and fluoride ion (F–) are, (a) 4 and 2, (b) 6 and 6, (c) 8 and 4, (d) 4 and 8 (NEET-II 2016), , 23. The intermetallic compound LiAg crystallizes in, cubic lattice in which both lithium and silver have, coordination number of eight. The crystal class is, (a) face-centred cube (b) simple cube, (c) body-centred cube (d) none of these. (1997), , 15. The ionic radii of A+ and B– ions are 0.98 × 10–10 m, and 1.81 × 10–10 m. The coordination number of, each ion in AB is, (a) 8, (b) 2, (NEET-I 2016), (c) 6, (d) 4, , 24. In the fluorite structure, the coordination number, of Ca 2+ ion is, (a) 4, (b) 6, (c) 8, (d) 3, (1993), , 16. The number of octahedral void(s) per atom present, in a cubic close-packed structure is, (a) 1, (b) 3, (c) 2, (d) 4, (2012), 17. Structure of a mixed oxide is cubic close packed, (ccp). The cubic unit cell of mixed oxide is composed, of oxide ions. One fourth of the tetrahedral voids, are occupied by divalent metal A and the octahedral, voids are occupied by a monovalent metal B. The, formula of the oxide is, (a) ABO2, (b) A2BO2, (c) A2B3O4, (d) AB2O2 (Mains 2012), 18. A solid compound XY has NaCl structure. If the radius, of the cation is 100 pm, the radius of the anion (Y–), will be, (a) 275.1 pm, (b) 322.5 pm, (c) 241.5 pm, (d) 165.7 pm(Mains 2011), 19. A compound formed by elements X and Y crystallises, in a cubic structure in which the X atoms are at the, corners of a cube and the Y atoms are at the facecentres. The formula of the compound is, (a) XY3, (b) X3Y, (c) XY, (d) XY2, (2004), 20. In cube of any crystal A-atom placed at every, corners and B-atom placed at every centre of face., The formula of compound is, (a) AB, (b) AB3, (c) A2B2, (d) A2B3, (2000), 21. In crystals of which one of the following ionic, compounds would you expect maximum distance, between centres of cations and anions?, (a) CsI, (b) CsF, (c) LiF, (d) LiI, (1998), 22. The second order Bragg diffraction of X-rays with, λ = 1.00 Å from a set of parallel planes in a metal, occurs at an angle 60°. The distance between the, scattering planes in the crystal is, (a) 2.00 Å, (b) 1.00 Å, (c) 0.575 Å, (d) 1.15 Å, (1998), , www.neetujee.com, , 1.7 Packing Efficiency, 25. An element has a body centered cubic (bcc) structure, with a cell edge of 288 pm. The atomic radius is, (a) 3  288 pm, (b) 2  288 pm, 4, 4, 4, 4, (c), (d)  288 pm,  288 pm, 2, 3, (NEET 2020), 26. The vacant space in bcc lattice unit cell is, (a) 48%, (b) 23%, (c) 32%, (d) 26% (2015, 2008), 27. If a is the length of the side of a cube, the distance, between the body-centred atom and one corner, atom in the cube will be, 2, 4, (a), a, (b) a, 3, 3, 3, 3, (c), a, (d), a, (2014), 4, , 2, , 28. A metal crystallises with a face-centred cubic lattice., The edge of the unit cell is 408 pm. The diameter of, the metal atom is, (a) 288 pm, (b) 408 pm, (c) 144 pm, (d) 204 pm, (2012), 29. AB crystallizes in a body-centred cubic lattice, with edge length ‘a’ equal to 387 pm. The distance, between two oppositely charged ions in the lattice is, (a) 335 pm, (b) 250 pm, (c) 200 pm, (d) 300 pm, (2010), 30. Lithium metal crystallises in a body-centred cubic, crystal. If the length of the side of the unit cell of, lithium is 351 pm, the atomic radius of lithium will, be, (a) 151.8 pm, (b) 75.5 pm, (c) 300.5 pm, (d) 240.8 pm, (2009), 31. Copper crystallises in a face-centred cubic lattice, with a unit cell length of 361 pm. What is the radius, of copper atom in pm?, (a) 157, (b) 181, (c) 108, (d) 128, (2009), , www.mediit.in

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The Solid State, 32. Which of the following statements is not correct?, (a) The number of carbon atoms in a unit cell of, diamond is 8., (b) The number of Bravais lattices in which a crystal, can be categorized is 14., (c) The fraction of the total volume occupied by the, atoms in a primitive cell is 0.48., (d) Molecular solids are generally volatile. (2008), 33. If a stands for the edge length of the cubic systems:, simple cubic, body-centred cubic and face-centred, cubic, then the ratio of radii of the spheres in these, systems will be respectively, 1, 3, 2, a:, (a) a :, a (b) 1a :, 3a : 2a, 2, 2, 2, 1, 1, 3, 1, 1, 3a :, a, a: a:, a (d) a :, (c), 2, 4, 2, 2 2, 2, (2008), 34. The fraction of total volume occupied by the atoms, present in a simple cube is, p, p, p, p, (a), (b), (d) 6, (c) 4, 32, 42, (2007), 35. The pyknometric density of sodium chloride crystal, is 2.165 × 103 kg m–3 while its X-ray density is, 2.178 × 103 kg m–3. The fraction of unoccupied sites, in sodium chloride crystal is, (a) 5.96, (b) 5.96 × 10–2, (c) 5.96 × 10–1, (d) 5.96 × 10–3, (2003), 36. The edge length of face-centred unit cubic cells is, 508 pm. If the radius of the cation is 110 pm, the, radius of the anion is, (a) 144 pm, (b) 398 pm, (1998), (c) 288 pm, (d) 618 pm, , 1.8 Calculations Involving Unit Cell Dimensions, 37. Iron exhibits bcc structure at room temperature., Above 900°C, it transforms to fcc structure. The, ratio of density of iron at room temperature to that, at 900°C (assuming molar mass and atomic radii of, iron remains constant with temperature) is, 1, 4, 3, (d) 2, (b) 3 3, (c) 4 3, (a) 3, 2, 2, 2, (NEET 2018), 38. Lithium has a bcc structure. Its density is, 530 kg m–3 and its atomic mass is 6.94 g mol–1., Calculate the edge length of a unit cell of lithium, metal. (NA = 6.02 × 1023 mol–1), (a) 527 pm, (b) 264 pm, (c) 154 pm, (d) 352 pm, (NEET-I 2016), , www.neetujee.com, , 3, , 39. A metal has a fcc lattice. The edge length of the unit, cell is 404 pm. The density of the metal is 2.72 g cm–3., The molar mass of the metal is, (NA Avogadro’s constant = 6.02 × 1023 mol–1), (a) 27 g mol–1, (b) 20 g mol–1, –1, (c) 40 g mol, (d) 30 g mol–1, (NEET 2013), 40. CsBr crystallises in a body-centred cubic lattice., The unit cell length is 436.6 pm. Given that the, atomic mass of Cs = 133 and that of Br = 80 amu, and Avogadro number being 6.02 × 1023 mol–1, the, density of CsBr is, (a) 4.25 g/cm3, (b) 42.5 g/cm3, 3, (c) 0.425 g/cm, (d) 8.25 g/cm3, (2006), 41. An element (atomic mass = 100 g/mol) having bcc, structure has unit cell edge 400 pm. The density of, element is, (a) 7.289 g/cm3, (b) 2.144 g/cm3, 3, (c) 10.376 g/cm, (d) 5.188 g/cm3, (1996), , 1.9 Imperfections in Solids, 42. Formula of nickel oxide with metal deficiency defect, in its crystal is Ni0.98O. The crystal contains Ni2+ and, Ni3+ ions. The fraction of nickel existing as Ni2+ ions, in the crystal is, (a) 0.96, (b) 0.04, (c) 0.50, (d) 0.3, (Odisha NEET 2019), 43. The correct statement regarding defects in crystalline, solids is, (a) Frenkel defects decrease the density of, crystalline solids, (b) Frenkel defect is a dislocation defect, (c) Frenkel defect is found in halides of alkaline, metals, (d) Schottky defects have no effect on the density of, crystalline solids., (2015), 44. The appearance of colour in solid alkali metal, halides is generally due to, (a) interstitial positions, (b) F-centres, (c) Schottky defect, (d) Frenkel defect., (2006), 45. Schottky defect in crystals is observed when, (a) density of the crystal is increased, (b) unequal number of cations and anions are, missing from the lattice, (c) an ion leaves its normal site and occupies an, interstitial site, (d) equal number of cations and anions are missing, from the lattice., (1998), , www.mediit.in

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4, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 46. Ionic solids, with Schottky defects, contain in their, structure, (a) cation vacancies only, (b) cation vacancies and interstitial cations, (c) equal number of cation and anion vacancies, (d) anion vacancies and interstitial anions., (1994), , semiconductor?, (a) Selenium, (c) Germanium, , 1.10 Electrical Properties, 47. Which is the incorrect statement?, (a) Density decreases in case of crystals with, Schottky defect., (b) NaCl(s) is insulator, silicon is semiconductor,, silver is conductor, quartz is piezoelectric, crystal., (c) Frenkel defect is favoured in those ionic, compounds in which sizes of cation and anions, are almost equal., (d) FeO0.98 has non-stoichiometric metal deficiency, defect., (NEET 2017), 48. With which one of the following elements, silicon should be doped so as to give p-type of, , (b) Boron, (d) Arsenic, , (2008), , –4, , 49. If NaCl is doped with 10 mol % of SrCl2,, the concentration of cation vacancies will be, (NA = 6.02 × 1023 mol–1), (a) 6.02 × 1016 mol–1, (b) 6.02 × 1017 mol–1, 14, –1, (c) 6.02 × 10 mol, (d) 6.02 × 1015 mol–1, (2007), 50. If we mix a pentavalent impurity in a crystal lattice of, germanium, what type of semiconductor formation, will occur?, (a) n-type semiconductor, (b) p-type semiconductor, (c) Both (a) and (b), (d) None of these, (1996), 51. On doping Ge metal with a little of In or Ga, one, gets, (a) p-type semiconductor, (b) n-type semiconductor, (c) insulator, (d) rectifier., (1993), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., , (b), (c), (a), (d), (d), (a), , 2., 12., 22., 32., 42., , (c), (c), (d), (c), (a), , 3., 13., 23., 33., 43., , (d), (d), (c), (c), (b), , 4., 14., 24., 34., 44., , (b), (c), (c), (d), (b), , 5., 15., 25., 35., 45., , (a), (c), (a), (d), (d), , 6., 16., 26., 36., 46., , (d), (a), (c), (a), (c), , 7., 17., 27., 37., 47., , (b) 8., (d) 18., (d) 28., (c) 38., (c, d) 48., , (d), (c), (a), (d), (b), , 9., 19., 29., 39., 49., , (c), (a), (a), (a), (b), , 10., 20., 30., 40., 50., , (c), (b), (a), (a), (a), , Hints & Explanations, 1. (b) : Liquid crystals on heating first become turbid, and then on further heating turbidity completely, disappears., 2. (c) : Glass is a supercooled liquid which forms a, non-crystalline solid without a regular lattice., 3. (d) : Crystals show good cleavage because their, constituent particles are arranged in planes., 4. (b) : The phenomenon of existence of a substance, in two or more crystalline structures is called, polymorphism., 5. (a) : The electrostatic force of attraction which, exists between oppositely charged ions is called as ionic, bond., , www.neetujee.com, , 6., , (d) : KI is an ionic compound while benzene is not., , 7., , (b) : For orthorhombic system, α = β = γ = 90º, , 8. (d) : Diamond is like ZnS (Zinc blende)., Carbon forming ccp (fcc) and also occupying half of, tetrahedral voids., Total no. of carbon atoms per unit cell, 1, 1, 8,  6, , 4, 8, 8, 2, (Corners), , (Face, centred), , (Tetrahedral, void), , 9. (c) : Here given unit cell is shared equally by six, faces in the fcc which is shared equally by six different, unit cells., , www.mediit.in

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The Solid State, , 5, , 10. (c) : hcp is a closed packed arrangement in which, the unit cell is hexagonal and coordination number is 12., 11. (c) : The contribution of eight atoms of face-centred, 1, cubic unit cell = 8 , = 1 atom. There is one atom at, 8, each of six faces, which is shared by 2 unit cells each. The, 1, contribution of 6 face-centred atoms = 6  = 3., 2, Therefore n = 1 + 3 = 4, 12. (c) : fcc crystal contains, 1, 1,  8   6   4 atoms in a unit cell, 8, , Number of A2+ ions  8  1  2, 4, Number of octahedral voids = Number of B+ ions = N = 4, Ratio, O2– : A2+ : B+ = 4 : 2 : 4 = 2 : 1 : 2, Formula of oxide = AB2O2, r, 18. (c) : For NaCl,   0.414, r , Given : radius of cation = 100 pm, 100, 100,  0.414 ,  r  r  241.5 pm, r, , 0.414, 19. (a) : In a unit cell, X atoms at the corners, 1, = 81, 8, 1, Y atoms at the face centres =  6  3, 2, Ratio of X and Y = 1 : 3. Hence formula is XY3., 20. (b) : ‘A’ atoms are at ‘8’ corners of the cube. Thus,, 1, no. of ‘A’ atoms per unit cell = 8 8  1, ‘B’ atoms are at the face centre of six faces. Thus, no. of ‘B’, 1, atoms per unit cell = 6   3, 2, The formula is AB ., , 2, , 3, , 21. (a) : As Cs+ ion has larger size than Li+ and I– has, larger size than F–, so maximum distance between, centres of cations and anions is in CsI., 13. (d) : Number of atoms per unit cell in hcp = 6, Number of octahedral void in hcp = 6, Number of anions per unit cell = 6, Number of octahedral voids occupied by cations, , , Formula of compound = C A = C A, 9/2, , 6, , 75 9, 6 , 100 2, , 34, , 14. (c) : In fluorite structure, Ca2+ ions are in the facecentred cubic arrangement. Each Ca2+ is connected to, 4 F– ions below it and to another set of 4 F– ions above it, i.e. Ca2+ has a coordination number of 8 and each F– ion, has a coordination number 4., 10, r 0.9810,  0.541, , 10, r 1.8110, It lies in the range of 0.414 to 0.732 hence, coordination, number of each ion will be 6 as the compound will have, NaCl type structure i.e., octahedral arrangement., , 15. (c) : Radius ratio,, , 16. (a) : Number of octahedral voids is same as number, of atoms., 17. (d) : Number of atoms in ccp = 4 = O2–, Number of tetrahedral voids = 2 × N = 2 × 4 = 8, , www.neetujee.com, , 22. (d) : According to Bragg’s equation, n = 2d sin , As, n = 2,  = 1.00Å,  = 60°, d = ?, 2dsin = n, 2dsin60° = 2 × 1 Å, , 3, 2, 3 , 2d ,  2 d, = 1.15 Å  sin 60 , 2, 2, , 3, 23. (c) : A body-centred cubic unit cell consists of 8, atoms at the corners and one atom at the centre., 24. (c) : In fluorite (CaF ) structure, C.N. of, 2, , Ca2+ = 8, C.N. of F– = 4., 3, , 25. (a) : For bcc structure, r  a, where a is the unit, 4, , cell edge length and r is the radius of the sphere (atom)., r, , 3,  288 pm, 4, , 26. (c) : Packing efficiency of bcc lattice = 68%, Hence, empty space = 32%., 27. (d) : The distance between the, body-centred atom and one corner, 3a, atom is i.e. half of the body, 2, diagonal., , www.mediit.in

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6, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 28. (a) : For a face-centred cubic (fcc) structure,, 408, a,  144 pm, r  , a = 408 pm, r , 2 2, , 2 2, , Diameter = 2r = 2 × 144 = 288 pm, 29. (a) : For a bcc lattice, 2(r+ + r–) = 3 a, where r+ = radius of cation, r– = radius of anion, a = edge length, , 3 387 = 335.15 pm ≈ 335 pm, 2, 30. (a) : Since Li crystallises in body-centred cubic, crystal, atomic radius,, 3a, r, (a  edge length), 4,  r  3  351151.8 pm (Given: a = 351 pm), 4, 31. (d) : Since Cu crystallises in a face-centred cubic, lattice,, a, Atomic radius, r , (a = edge length = 361 pm), 2 2, 361, r,  127.6  128 pm, 2 2,  (r, , , ,  r ) , , 32. (c) : Packing fraction for a cubic unit cell is given by, 4, Z  r3, f 3 3, a, where a = edge length, r = radius of cation and anion., Efficiency of packing in simple cubic or primitive cell, = π/6 = 0.52 i.e. 52% of unit cell is occupied by atoms and, 48% is empty., 33. (c) : For simple cubic : r = a/2, For body centred : r  a 3 / 4, For face-centred: r , , a, , 2 2, where a = edge length, r = radius., a a,  Ratio of radii of the three will be : 3 a, :, 2 4 2 2, 34. (d), 35. (d) : Molar volume from pyknometric density, =, , M, , m3, , 2.165  10, M, Molar volume from X-ray density =, m3, 3, M  1 1  3 2.178  10, , m, Volume occupied =, , , 103  2.165 2.178  , , www.neetujee.com, , 3, , , , , , Fraction unoccupied,  0.013 M 103   M 103 , , = ,  = 5.96 × 10–3, , , , , ,  2.165  2.178 , 2.165 , 36. (a) : In the face–centred cubic lattice, the edge, length of the unit cell, a = r + 2R + r, where r = Radius of cation, R = Radius of anion,  508 = 2 × 110 + 2R  R = 144 pm, , , , , 37. (c) : For bcc lattice : Z = 2, a  4r, 3, For fcc lattice : Z = 4, a  2 2 r, , ,  ZM3 , N a,  A  bcc, dR.T., , , d900 C  ZM , , , N a3,  A  fcc, Given, molar mass and atomic radii are constant., 3, 3, 2  2 2r , 3,   , 4 4r, 2, , 4,  3 , 38. (d) : For bcc, Z = 2, ρ = 530 kg m–3,, at. mass of Li = 6.94 g mol–1, NA = 6.02 × 1023 mol–1, 530 1000 g, ρ = 530 kg m–3 =, = 0.53 g cm–3, 3, 1  (100) cm3, Z  At. mass, , NA  a3, Z  At. mass 2  6.94, a3 , , 6.02 1023  0.53, NA  , –24, = 43.5 × 10 cm3,  a = 352 × 10–10 cm = 352 pm, ZM, 39. (a) : d , NAa3 (Z = 4 for fcc), 3, , , , 2.72  6.023 1023  404 1010, d  NA  a3, , M, 4, Z, M = 27 g mol–1, ZM, 40. (a) : Density of CsBr =, , , , a3  NA, 1  213, 3, ,  4.25 g/cm, 10 3, (436.6 10 )  6.023 1023, (It has one formula unit in the unit cell, so Z = 1.), 41. (d) : Cell edge = 400 pm; number of atoms in bcc (Z), = 2 and atomic mass = 100 g/mol., Since atomic mass is 100 g/mol, therefore mass, 100, of each atom (m) =, = 16.6  10–23 g, 6.023  1023, , www.mediit.in

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The Solid State, , 7, , We know that volume of unit cell = (400 pm)3, = (64  106)pm3 = 64  10−24 cm3 and, mass of unit cell = Z  m = 2  (16.6  10–23), = 33.2  10–23 g, Mass of unit cell, Therefore density =, Volume of unit cell, 33.2  1023, =, = 5.188 g/cm3, 64  1024, , 47. (c, d) : Frenkel defect is favoured in those ionic, compounds in which there is large difference in the size, of cations and anions., Non-stoichiometric defects due to metal deficiency is, shown by FexO where x = 0.93 to 0.96., 48. (b) : If silicon is doped with any of the elements of, group 13 (B, Al, Ga, In, Tl) of the periodic table, p-type, of semiconductor will be obtained., , 42. (a) : Let the fraction of metal which exists as Ni2+, ion be x. Then the fraction of metal as Ni3+ = 0.98 – x,  2x + 3(0.98 – x) = 2,  2x + 2.94 – 3x = 2  x = 0.94, 43. (b) : Frenkel defect is a dislocation defect as smaller, ions (usually cations) are dislocated from normal sites to, interstitial sites. Frenkel defect is shown by compounds, having large difference in the size of cations and anions, hence, alkali metal halides do not show Frenkel defect., Also, Schottky defect decreases the density of crystal, while Frenkel defect has no effect on the density of, crystal., 44. (b) : F-centres are the sites where anions are missing, and instead electrons are present. They are responsible, for colours., 45. (d) : In Schottky defect, equal no. of cations and, anions are missing from the lattice. So, the crystal, remains neutral. Such defect is more common in highly, ionic compounds of similar cationic and anionic size, i.e., NaCl., 46. (c) : When an atom is missing from its normal lattice, site, a lattice vacancy is created. Such a defect, which, involves equal number of cation and anion vacancies in, the crystal lattice is called a Schottky defect., , 49. (b) : As each Sr2+ ion introduces one cation vacancy,, therefore, concentration of cation vacancies = mole % of, SrCl2 added.,  Concentration of cation vacancies = 10–4 mole%, , , , 104,  6.023 1023  6.023 1017, 100, , 50. (a) : When an impurity atom with 5 valence electrons, (as arsenic) is introduced in a germanium crystal, it, replaces one of the germanium atoms. Four of the five, valence electrons of the impurity atom form covalent, bonds with each valence electron of four germanium, atoms and fifth valence electron becomes free to move, in the crystal structure. This free electron acts as a charge, carrier. Such as an impure germanium crystal is called, n-type semiconductor because in it charge carriers are, negative (free electrons)., 51. (a) : p-type of semiconductors are produced, (a) due to metal deficiency defects (b) by adding impurity, containing less electrons (i.e. atoms of group 13)., Ge belongs to Group 14 and In or Ga to Group 13., Hence on doping p-type semiconductor is obtained. This, doping of Ge with In increase the electrical conductivity, of the Ge crystal., , , , , , , , , , , , , , , , , , , , , , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 2, , 2.2, 1., , Expressing Concentration of Solutions, , Which of the following is dependent on temperature?, (a) Molarity, (b) Mole fraction, (c) Weight percentage, (d) Molality, (NEET 2017), , 2., , What is the mole fraction of the solute in a, 1.00 m aqueous solution?, (a) 1.770, (b) 0.0354, (c) 0.0177, (d) 0.177, (2015, 2011), , 3., , How many grams of concentrated nitric acid, solution should be used to prepare 250 mL of, 2.0 M HNO3? The concentrated acid is 70% HNO3., (a) 70.0 g conc. HNO3, (b) 54.0 g conc. HNO3, (c) 45.0 g conc. HNO3, (d) 90.0 g conc. HNO3, (NEET 2013), , 4., , 5., , Solutions, 8., , How many g of dibasic acid (mol. weight 200), should be present in 100 mL of the aqueous solution, to give strength of 0.1 N?, (a) 10 g, (b) 2 g, (c) 1 g, (d) 20 g, (1999), , 9., , What is the molarity of H2SO4 solution, that has, a density 1.84 g/cc at 35°C and contains 98% by, weight?, (a) 18.4 M, (b) 18 M, (c) 4.18 M, (d) 8.14 M, (1996), , 10., , The concentration unit, independent of temperature,, would be, (a) normality, (b) weight volume percent, (c) molality, (d) molarity. (1995, 1992), , 11., , How many grams of CH3OH should be added to, water to prepare 150 mL solution of 2 M CH3OH?, (a) 9.6 × 103, (b) 2.4 × 103, (c) 9.6, (d) 2.4, (1994), , Which of the following compounds can be used as, antifreeze in automobile radiators?, (a) Methyl alcohol, (b) Glycol, (c) Nitrophenol, (d) Ethyl alcohol, (2012), , 12., , Concentrated aqueous sulphuric acid is 98% H2SO4, by mass and has a density of 1.80 g mL–1. Volume, of acid required to make one litre of 0.1 M H2SO4, solution is, (a) 16.65 mL, (b) 22.20 mL, (c) 5.55 mL, (d) 11.10 mL (2007), , In water saturated air, the mole fraction of water, vapour is 0.02. If the total pressure of the saturated, air is 1.2 atm, the partial pressure of dry air is, (a) 1.18 atm, (b) 1.76 atm, (c) 1.176 atm, (d) 0.98 atm., (Odisha NEET 2019), , 13., , pA and pB are the vapour pressures of pure liquid, components, A and B, respectively of an ideal, binary solution. If xA represents the mole fraction of, component A, the total pressure of the solution will, be, (a) pA + xA(pB – pA), (b) pA + xA ( pA – pB), (c) pB + xA(pB – pA), (d) pB + xA ( pA – pB), (2012), Vapour pressure of chloroform (CHCl 3) and, dichloromethane (CH2Cl2) at 25°C are 200 mm Hg, and 41.5 mm Hg respectively. Vapour pressure of, the solution obtained by mixing 25.5 g of CHCl3 and, 40 g of CH2Cl2 at the same temperature will be, , 6., , The mole fraction of the solute in one molal aqueous, solution is, (a) 0.009, (b) 0.018, (c) 0.027, (d) 0.036, (2005), , 7., , 2.5 litre of 1 M NaOH solution is mixed with, another 3 litre of 0.5 M NaOH solution. Then find, out molarity of resultant solution., (a) 0.80 M, (b) 1.0 M, (c) 0.73 M, (d) 0.50 M, (2002), , www.neetujee.com, , 2.4, , 14., , Vapour Pressure of Liquid Solutions, , www.mediit.in

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Solutions, , 9, , (Molecular mass of CHCl3 = 119.5 u and molecular, mass of CH2Cl2 = 85 u), (a) 173.9 mm Hg, (b) 615.0 mm Hg, (c) 347.9 mm Hg, (d) 285.5 mm Hg, (Mains 2012), 15., , A solution has a 1 : 4 mole ratio of pentane to hexane., The vapour pressures of the pure hydrocarbons at, 20 °C are 440 mm Hg for pentane and 120 mm Hg for, hexane. The mole fraction of pentane in the vapour, phase would be, (a) 0.200, (b) 0.549, (c) 0.786, (d) 0.478, (2005), , 16., , The vapour pressure of two liquids P and Q are 80, and 60 torr, respectively. The total vapour pressure, of solution obtained by mixing 3 mole of P and, 2 mol of Q would be, (a) 72 torr, (b) 140 torr, (c) 68 torr, (d) 20 torr, (2005), , 2.5, 17., , 18., , Ideal and Non-Ideal Solutions, , The mixture which shows positive deviation from, Raoult’s law is, (a) ethanol + acetone (b) benzene + toluene, (c) acetone + chloroform, (d) chloroethane + bromoethane., (NEET 2020), For an ideal solution, the correct option is, (a) mixG = 0 at constant T and P, (b) mixS = 0 at constant T and P, (c)  V  0 at constant T and P, (d) mix, mix H = 0 at constant T and P., , (NEET 2019), , 19. The mixture that forms maximum boiling azeotrope, is, (a) heptane + octane, (b) water + nitric acid, (c) ethanol + water, (d) acetone + carbon disulphide., (NEET 2019), 20. Which of the following statements is correct, regarding a solution of two components A and B, exhibiting positive deviation from ideal behaviour?, (a) Intermolecular attractive forces between A-A, and B-B are stronger than those between A-B., (b) ∆mix H = 0 at constant T and P., (c) ∆mix V = 0 at constant T and P., (d) Intermolecular attractive forces between, A-A and B-B are equal to those between A-B., (Odisha NEET 2019), 21. Which one of the following is incorrect for ideal, solution?, (a) Hmix = 0, (b) Umix = 0, (c) P = Pobs – Pcalculated by Raoult’s law = 0, (d) Gmix = 0, (NEET-II 2016), , www.neetujee.com, , 22. Which of the following statements about the, composition of the vapour over an ideal 1 : 1, molar mixture of benzene and toluene is correct?, Assume that the temperature is constant at 25°C., (Given, vapour pressure data at 25°C, benzene =, 12.8 kPa, toluene = 3.85 kPa), (a) The vapour will contain equal amounts of, benzene and toluene., (b) Not enough information is given to make a, prediction., (c) The vapour will contain a higher percentage of, benzene., (d) The vapour will contain a higher percentage of, toluene., (NEET-I 2016), 23. Which condition is not satisfied by an ideal solution?, (a) mixV = 0, (b) mixS = 0, (c) Obeyance to Raoult’s Law, (d) mixH = 0, (Karnataka NEET 2013), 24. A solution of acetone in ethanol, (a) obeys Raoult’s law, (b) shows a negative deviation from Raoult’s law, (c) shows a positive deviation from Raoult’s law, (d) behaves like a near ideal solution., (2006), 25. A solution containing components A and B follows, Raoult’s law, (a) A - B attraction force is greater than A - A and, (b) B - B, A - B attraction force is less than A - A and B - B, (c) A - B attraction force remains same as A - A and, B-B, (d) volume of solution is different from sum of, volume of solute and solvent., (2002), 26. All form ideal solution except, (a) C6H6 and C6H5CH3 (b) C2H6 and C2H5I, (c) C6H5Cl and C6H5Br (d) C2H5I and C2H5OH, (1988), 27. An ideal solution is formed when its components, (a) have no volume change on mixing, (b) have no enthalpy change on mixing, (c) have both the above characteristics, (d) have high solubility., (1988), , 2.6, , Colligative Properties and, Determination of Molar Mass, , 28. The freezing point depression constant (Kf) of, benzene is 5.12 K kg mol–1. The freezing point, depression for the solution of molality 0.078 m, containing a non-electrolyte solute in benzene is, (rounded off upto two decimal places), (a) 0.20 K, (b) 0.80 K, (c) 0.40 K, (d) 0.60 K (NEET 2020), , www.mediit.in

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10, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 29. If molality of the dilute solution is doubled, the, value of molal depression constant (Kf) will be, (a) halved, (b) tripled, (c) unchanged, (d) doubled. (NEET 2017), 30. At 100°C the vapour pressure of a solution of 6.5 g, of a solute in 100 g water is 732 mm. If Kb = 0.52, the, boiling point of this solution will be, (a) 102 °C, (b) 103 °C, (c) 101 °C, (d) 100 °C, (NEET-I 2016), 31. 200 mL of an aqueous solution of a protein contains, its 1.26 g. The osmotic pressure of this solution at, 300 K is found to be 2.57 × 10–3 bar. The molar mass, of protein will be (R = 0.083 L bar mol–1 K–1), (a) 51022 g mol–1, (b) 122044 g mol–1, –1, (c) 31011 g mol, (d) 61038 g mol–1, (Mains 2011), 32. A solution of sucrose (molar mass = 342 g mol–1), has been prepared by dissolving 68.5 g of sucrose in, 1000 g of water. The freezing point of the solution, obtained will be (Kf for water = 1.86 K kg mol–1), (a) – 0.372 °C, (b) – 0.520 °C, (c) + 0.372 °C, (d) – 0.570 °C (2010), 33. During osmosis, flow of water through a, semipermeable membrane is, (a) from solution having lower concentration only, (b) from solution having higher concentration only, (c) from both sides of semipermeable membrane, with equal flow rates, (d) from both sides of semipermeable membrane, with unequal flow rates., (2006), 34. 1.00 g of a non-electrolyte solute (molar mass, 250 g mol–1) was dissolved in 51.2 g of benzene., If the freezing point constant, Kf of benzene is, 5.12 K kg mol–1, the freezing point of benzene will, be lowered by, (a) 0.2 K, (b) 0.4 K, (c) 0.3 K, (d) 0.5 K, (2006), 3, , 35. A solution containing 10 g per dm of urea, (molecular mass = 60 g mol–1) is isotonic with a, 5% solution of a non-volatile solute. The molecular, mass of this non-volatile solute is, (a) 200 g mol–1, (b) 250 g mol–1, –1, (c) 300 g mol, (d) 350 g mol–1, (2006), 36. A solution of urea (mol. mass 56 g mol–1) boils at, 100.18°C at the atmospheric pressure. If Kf and Kb, for water are 1.86 and 0.512 K kg mol–1 respectively,, the above solution will freeze at, (a) 0.654°C, (b) – 0.654°C, (c) 6.54°C, (d) – 6.54°C, (2005), , www.neetujee.com, , 37. A solution contains non-volatile solute of molecular, mass M2. Which of the following can be used to, calculate the molecular mass of solute in terms of, osmotic pressure?,  m2 , VRT (b) M  m2  RT, (a) M , =,   , 2 , 2, , ,  m, V, , 2, RT, (c) M , (d) M   m2  , , RT, 2  , 2 , V, V, (m2 = mass of solute, V = volume of solution,,  = osmotic pressure), (2002), 38. Pure water can be obtained from sea water by, (a) centrifugation, (b) plasmolysis, (c) reverse osmosis, (d) sedimentation., (2001), 39. From the colligative properties of solution, which, one is the best method for the determination of, molecular weight of proteins and polymers?, (a) Osmotic pressure, (b) Lowering in vapour pressure, (c) Lowering in freezing point, (d) Elevation in boiling point, (2000), 40. The vapour pressure of benzene at a certain, temperature is 640 mm of Hg. A non-volatile and, non-electrolyte solid, weighing 2.175 g is added, to 39.08 of benzene. The vapour pressure of the, solution is 600 mm of Hg. What is the molecular, weight of solid substance?, (a) 69.5, (b) 59.6, (c) 49.50, (d) 79.8, (1999), 41. If 0.15 g of a solute, dissolved in 15 g of solvent, is, boiled at a temperature higher by 0.216°C, than that, of the pure solvent. The molecular weight of the, substance (Molal elevation constant for the solvent, is 2.16°C) is, (a) 10.1, (b) 100, (c) 1.01, (d) 1000, (1999), 42. A 5% solution of cane sugar (mol. wt. = 342), is isotonic with 1% solution of a substance X. The, molecular weight of X is, (a) 68.4, (b) 171.2, (c) 34.2, (d) 136.8, (1998), 43. The vapour pressure of a solvent decreased by 10 mm, of mercury when a non-volatile solute was added to, the solvent. The mole fraction of the solute in the, solution is 0.2. What should be the mole fraction of, the solvent if the decrease in the vapour pressure is, to be 20 mm of mercury?, (a) 0.4, (b) 0.6, (c) 0.8, (d) 0.2, (1998), , www.mediit.in

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Solutions, 44. The vapour pressure of CCl4 at 25°C is 143 mm Hg., If 0.5 g of a non-volatile solute (mol. weight = 65) is, dissolved in 100 g CCl4, the vapour pressure of the, solution will be, (a) 199.34 mm Hg (b) 143.99 mm Hg, (c) 141.43 mm Hg (d) 94.39 mm Hg. (1996), 45. The relationship between osmotic pressure at, 273 K when 10 g glucose (p1), 10 g urea (p2), and 10 g, sucrose (p3) are dissolved in 250 mL of water is, (a) p2 > p1 > p3, (b) p2 > p3 > p1, (c) p1 > p2 > p3, (d) p3 > p1 > p2, (1996), 46. According to Raoult’s law, the relative lowering of, vapour pressure for a solution is equal to, (a) mole fraction of solute, (b) mole fraction of solvent, (c) moles of solute, (d) moles of solvent., (1995), 47. If 0.1 M solution of glucose and 0.1 M solution of, urea are placed on two sides of the semipermeable, membrane to equal heights, then it will be correct to, say that, (a) there will be no net movement across the, membrane, (b) glucose will flow towards glucose solution, (c) urea will flow towards glucose solution, (d) water will flow from urea solution to glucose., (1992), 48. Which one is a colligative property?, (a) Boiling point, (b) Vapour pressure, (c) Osmotic pressure (d) Freezing point (1992), 49. Blood cells retain their normal shape in solution, which are, (a) hypotonic to blood (b) isotonic to blood, (c) hypertonic to blood (d) equinormal to blood., (1991), 50. The relative lowering of the vapour pressure is equal, to the ratio between the number of, (a) solute molecules to the solvent molecules, (b) solute molecules to the total molecules in the, solution, (c) solvent molecules to the total molecules in the, solution, (d) solvent molecules to the total number of ions of, the solute., (1991), , 2.7, , Abnormal Molar Masses, , 51. The van’t Hoff factor (i) for a dilute aqueous solution, of the strong electrolyte barium hydroxide is, (a) 0, (b) 1, (c) 2, (d) 3, (NEET-II 2016), , www.neetujee.com, , 11, , 52. The boiling point of 0.2 mol kg–1 solution of X in, water is greater than equimolal solution of Y in, water. Which one of the following statements is true, in this case?, (a) Molecular mass of X is less than the molecular, mass of Y., (b) Y is undergoing dissociation in water while X, undergoes no change., (c) X is undergoing dissociation in water., (d) Molecular mass of X is greater than the molecular, mass of Y., (2015, Cancelled), 53. Of the following 0.10 m aqueous solutions, which, one will exhibit the largest freezing point depression?, (a) KCl, (b) C6H12O6, (c) Al2(SO4)3, (d) K2SO4, (2014), 54. The van’t Hoff factor i for a compound which, undergoes dissociation in one solvent and, association in other solvent is respectively, (a) less than one and greater than one, (b) less than one and less than one, (c) greater than one and less than one, (d) greater than one and greater than one. (2011), 55. The freezing point depression constant for water is, –1.86 °C m–1. If 5.00 g Na2SO4 is dissolved in 45.0 g, H2O, the freezing point is changed by –3.82 °C., Calculate the van’t Hoff factor for Na2SO4., (a) 2.05, (b) 2.63, (c) 3.11, (d) 0.381, (2011), 56. A 0.1 molal aqueous solution of a weak, acid is 30% ionized. If Kf for water is 1.86°C/m, the, freezing point of the solution will be, (a) – 0.18 °C, (b) – 0.54 °C, (d) – 0.24 °C, (c) – 0.36 °C, (Mains 2011), 57. An aqueous solution is 1.00 molal in KI. Which, change will cause the vapour pressure of the solution, to increase?, (a) Addition of NaCl (b) Addition of Na2SO4, (c) Addition of 1.00 molal KI, (d) Addition of water, (2010), 58. A 0.0020 m aqueous solution of an ionic compound, [Co(NH3)5(NO2)]Cl freezes at –0.00732 °C. Number, of moles of ions which 1 mol of ionic compound, produces on being dissolved in water will be, (Kf = –1.86 °C/m), (a) 3, (b) 4, (c) 1, (d) 2 (2009), 59. 0.5 molal aqueous solution of a weak acid (HX) is, 20% ionised. If Kf for water is 1.86 K kg mol–1, the, lowering in freezing point of the solution is, (a) 0.56 K, (b) 1.12 K, (c) –0.56 K, (d) –1.12 K, (2007), , www.mediit.in

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12, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 60. Which of the following 0.10 m aqueous solution will, have the lowest freezing point?, (a) KI, (b) C12H22O11, (c) Al2(SO4)3, (d) C5H10O5, (1997), , 62. At 25°C, the highest osmotic pressure is exhibited, by 0.1 M solution of, (a) glucose, (b) urea, (c) CaCl2, (d) KCl., (1994), , 61. Which of the following salts has the same value of, van’t Hoff factor (i) as that of K3[Fe(CN)6]?, (a) Na2SO4, (b) Al(NO3)3, (c) Al2(SO4)3, (d) NaCl, (1994), , 63. Which of the following aqueous solution has, minimum freezing point?, (a) 0.01 m NaCl, (b) 0.005 m C2H5OH, (c) 0.005 m MgI2, (d) 0.005 m MgSO4 (1991), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., 61., , (a), (c), (d), (d), (b), (d), (b), , 2., 12., 22., 32., 42., 52., 62., , (c), (c), (c), (a), (a), (c), (c), , 3., 13., 23., 33., 43., 53., 63., , (c), (d), (b), (a), (b), (c), (a), , 4., 14., 24., 34., 44., 54., , (b) 5., None 15., (c) 25., (b) 35., (c) 45., (c) 55., , (c), (d), (c), (c), (a), (b), , 6., 16., 26., 36., 46., 56., , (b), (a), (d), (b), (a), (d), , 7., 17., 27., 37., 47., 57., , (c), (a), (c), (b), (a), (d), , 8., 18., 28., 38., 48., 58., , (c), (d), (c), (c), (c), (d), , 9., (a), 19. (b), 29. (c), 39. (a), 49. (b), 59. (b), , 10., 20., 30., 40., 50., 60., , (c), (a), (c), (a), (b), (c), , Hints & Explanations, 1. (a) : Molarity is a function of temperature as volume, depends on temperarure., 2. (c) : 1 molal aqueous solution means 1 mole of, solute is present in 1000 g of water., 1 1,  x, ,   0.0177, solute, 1000 56.5, 1, 18, wHNO3 1000, 3. (c) : Molarity  M, HNO3  Vsol(mL), wHNO3, or 2 , , 63, , 63,  1000, 250  wHNO3 = g, 2, , 70 63, , 100 2, Mass of acid = 45 g, 4. (b) : A 35% (V/V) solution of ethylene glycol is used, as an antifreeze in cars for cooling the engine. At this, concentration, the antifreeze lowers the freezing point of, water to 255.4 K (–17.6 °C)., 5. (c) : H2SO4 is 98% by weight., Weight of H2SO4 = 98 g,mass, Weight 100, of solution = 100 g,  Volume of solution =, , mL, density 1.80, = 55.55 mL = 0.0555 L, Mass of acid ×, , Molarity of solution =, , www.neetujee.com, , 98, 98  0.0555 18.02 M, , Let V mL of this H2SO4 is used to prepare, 1 litre of 0.1 M H2SO4.,  mM of concentrated H2SO4 = mM of dilute H2SO4, or, V × 18.02 = 1000 × 0.1, 1000  0.1, V,  5.55 mL, 18.02, 6. (b) : 1 molal aqueous solution means 1 mole of, solute present in 1 kg of H2O., 1000, 1 mole of solute present in, mole of H 2O, 18, 1, 18, , xsolute , 1018  0.01768  0 .018, 1000, 1, 18, 7. (c) : Molecular weight of NaOH = 40 g mol–1, 2.5 litre of 1 M NaOH solution contain, 40 g mol–1 × 1 mol L–1 × 2.5 L, = 40 × 2.5 g of NaOH, 3 litre of 0.5 M NaOH solution contain, 40 g mol–1 × 0.5 mol L–1 × 3 L, = 40 × 0.5 × 3 g of NaOH, If these two solutions are mixed, the volume of the, resultant solution = (2.5 + 3) = 5.5 litre., 5.5 litre of the resultant solution contain, = 40(2.5 + 1.5) g of NaOH, 1 litre of the resultant solution contain, 40  4, 40  4 g of NaOH =, mole of NaOH, 5.5, 5.5 , 40, The molarity of the resultant solution = 0.727  0.73 M, , www.mediit.in

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Solutions, 8. (c) : The strength of the solution is 0.1 N., 200, w VN, (Equivalent weight = 2 = 100), , E 1000, 100  0.1 100, 1g, 1000, 9. (a) : We know that 98% H2SO4 by weight means, 98 g of H2SO4 is present in 100 g of solution., Therefore, its weight is 98 and moles of H2SO4, Weight of H2SO4 98,  1, , Molecular weight 98, Mass, and volume of solution =, Density, 100, 54.35, ,  54.35 mL , L, 1.84, 1000, Therefore, molarity of H2SO4, Moles of H2SO4, 1 1000, , ,  18.4 M, Volume (in litres), 54.35, 10. (c) : The molality involves weights of the solute and, the solvent. Since the weight does not change with the, temperature, therefore molality does not depend upon, the temperature., 11. (c) : Since the molecular mass of CH3OH is 32,, therefore quantity of2CH, 150 mL solution,  3OH to prepare, 150  32 = 9.6 g, of 2 M CH3OH ,  1000 , 12. (c) : pwater vapour = xwater vapour × Ptotal, = 0.02 × 1.2 = 0.024 atm, Ptotal = pwater vapour + pdry air, 1.2 = 0.024 + pdry air, pdry air = 1.2 – 0.024 = 1.176 atm, Partial vapour pressure is directly proportional to mole, fraction, p  x., 13. (d) : According to Raoult’s law,, P = xApA + xBpB, ... (i), For binary solutions,, xA + xB = 1, xB = 1 – xA, ... (ii), Putting value of xB from eqn. (ii) to eqn. (i), P = xApA + (1 – xA)pB = xApA + pB – xApB, P = pB + xA(pA – pB), 14. (None) :, p°CHCl3 = 200 mm Hg, p°CH2Cl2 = 41.5 mm Hg, Moles of CHCl3, , 13, , Ptotal = p°CHCl3 xCHCl3 + p°CH2Cl2 xCH2Cl2, = 200 × 0.31, ×0.69 = 62 + 28.63 = 90.63 mm Hg, nC H+ 41.5, 5 12 1, , 15. (d) :, nC, , w, , Weight, 25.5, , ,  0.213, Molecular weight, 119.5, 40, Moles of CH Cl   0.470, 2, 2, 85, 0.213,  0.31, xCHCl3 , 0.213  0.470, 0.470,  0.69, xCH2Cl2 , 0.213  0.470, , www.neetujee.com, , 6H14, , 4, , 1, 4, and xC H , 6, 14, 5, 5, pC 5H12  440 mm Hg; pC 6 H14  120 mm Hg, Ptotal  pC5 H12 xC5H12  pC6H14 xC6H14, ,  xC5H12 , , 1, 4,  440  120   88  96  184 mm of Hg, 5, 5, By Raoult’s law, pC H  pC H xC H, 5 12, 5 12, 5 12, x, , mole fraction of pentane in solution., C5H12, By Dalton’s law, pC5H12  xC5H12 Ptotal, xC, , 5H12, , ...(1), , ...(2), ,  mole fraction of pentane above the solution., , From (1) and (2),, 1, pC H  440  = 88 mm of Hg, 5 12, 5,  88  xC5H12 184, 88,  0.478, 184, 16. (a) : By Raoult’s Law, PT = pP° x°P + pQ° xQ, 3, 2, where pP° = 80 torr, pQ° = 60 torr, x P  5; xQ  5, 3, 2, P  80   60 , = 48 + 24 = 72 torr, T, 5, 5, 17. (a) : Mixture of ethanol and acetone shows positive, deviation from Raoult’s law., In pure ethanol, molecules are hydrogen bonded. On, adding acetone, its molecules get in between the host, molecules and break some of the hydrogen bonds, between them. Due to weakening of interactions, the, solution shows positive deviation from Raoult’s law., xC, , 5H12, , =, , 18. (d) : For an ideal solution, mixH = 0 and mixV = 0, at constant T and P., 19. (b) : Maximum boiling azeotropes are formed by, those solutions which show negative deviations from, Raoult’s law. H2O and HNO3 mixture shows negative, deviations., 20. (a) : In case of positive deviation from Raoult’s law,, A-B interactions are weaker than those between A-A or, B-B, i.e., in this case the intermolecular attractive forces, between the solute-solvent molecules are weaker than, those between the solute-solute and solvent-solvent, molecules. This means that in such solution, molecules, of A (or B) will find it easier to escape than in pure state., This will increase the vapour pressure and result in, positive deviation., , www.mediit.in

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14, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 21. (d) : For an ideal solution,, Hmix = 0, Vmix = 0,, Now, Umix = Hmix – PVmix,  Umix = 0, Also, for an ideal, solution,, pA  x p , p  x p, A A, , B, , B B, ,  P = Pobs – Pcalculated by Raoult’s law = 0, Gmix = Hmix – TSmix, For an ideal solution, Smix  0,  Gmix  0, 22. (c) : pBenzene = xBenzene p°Benzene, pToluene = xToluene p°Toluene, For an ideal11 : 1 molar mixture, 1 of benzene and toluene,, xBenzene  and xToluene , 2, 2, 1, 1, pBenzene  pBenzene  12.8 kPa = 6.4 kPa, 2, 2, 1, 1, pToluene  pToluene   3.85 kPa =1.925 kPa, 2, 2, Thus, the vapour will contain a high percentage of, benzene as the partial vapour pressure of benzene is, higher as compared to that of toluene., 23. (b) : For an ideal solution :, • Volume change (V) on mixing should be zero., • Heat change (H) on mixing should be zero., • Obeys Raoult’s law at every range of concentration., • Entropy change (S) on mixing  0., 24. (c) : Both the components escape easily showing, higher vapour pressure than the expected value. This, is due to breaking of some hydrogen bonds between, ethanol molecules., 25. (c) : Raoult’s law is valid for ideal solutions only., The element of non-ideality enters into the picture, when the molecules of the solute and solvent affect each, others intermolecular forces. A solution containing, components of A and B behaves as ideal solution when, A – B attraction force remains same as A – A and B – B, attraction forces., 26. (d) : Because C2H5I and C2H5OH are dissimilar, liquids., 27. (c) : For ideal solution,, Vmixing = 0 and Hmixing = 0., 28. (c) : Given : Kf = 5.12 K kg mol–1, m = 0.078 m, Tf = Kf × m = 5.12 × 0.078 = 0.39936 ≈ 0.40 K, 29. (c) : The value of molal depression constant, Kf is, constant for a particular solvent, thus, it will be unchanged, when molality of the dilute solution is doubled., 30. (c) : Given : WB = 6.5 g, WA = 100 g,, ps = 732 mm, Kb = 0.52, T °b= 100°C, p° = 760 mm, , www.neetujee.com, , p ps n2, n2, 760  732, p  n1 , 760  100 /18, 28 100,  n2  760 18  0.2046 mol, T b = K b × m, n2 1000, Tb – T °b= Kb , WA (g), 0.52  0.2046 1000, Tb100C ,  1.06, 100, Tb = 100 + 1.06 = 101.06 °C, w, 31. (d) : We know that pV = nRT, where n , M, w, V  RT, M, wRT 1.26  0.083  300, M, , 200, 2.57 103 , V, 1000, 1.26  0.083  300, , = 61038 g mol–1, 2.57 103  0.2, 32. (a) : We know, Tf = Kf m, w, m  B 1000 68.5 1000 68.5, , , , MB WA 342 1000 342, 68.5, T  1.86 ,  0.372C, f, 342,  Tf = 0 – 0.372°C = – 0.372°C, 33. (a), 1000  K f  wB, 34. (b) : MB , WA  Tf, 1000  5.12 1, or, 250 , 51.2  Tf, 1000  5.12 1,  0.4 K,  Tf , 51.2  250, 10, 3, 35. (c) : Molar concentration of urea =, dm, 60, Molar concentration of non-volatile solution, 50 1 50, L =, dm3, =, MB, MB, 10 50, For isotonic solutions,, , 60 M B,  MB = 300 g mol–1, ...(1), 36. (b) : Tf = Kf m, Tb = Kb m, ...(2), , , Tf, , , , Kf, Kb, , Tb, Tf  depression in freezing point, Tb  elevation in boiling point, , ...(3), , www.mediit.in

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Solutions, , 15, , Kf = 1.86 K kg mol–1, Kb = 0.512 K kg mol–1, Tb = 100.18 – 100 = 0.18, Tf 1.86,  From eq. (3),, , 0.18 0.512,  Tf = 0.654 = T°f – Tf = 0 – Tf  Tf = – 0.654°C,  Freezing point of urea in water = –0.654°C, n, 37. (b) : For dilute solution,  VRT,  V , , m2, , RT, M2, ,  M2 , , m2RT, , V, 38. (c), 39. (a), 40. (a) : p° = 640 mm Hg, ps = 600 mm Hg,, wB = 2.175 g, WA = 39.08 g, From Raoult’s law, p ps wB  MA 640  600 2.175  78, , , , 640, p, WA  M B, 39.08  M B,  MB = 69.5, 41. (b) : wB = 0.15 g, WA = 15 g, Tb = 0.216°C, Kb = 2.16, m = ?, As Tb =,  M B, , 1000  Kb  wB, MB  WA, , 1000  2.16  0.15,  100, 0.216 15, , 42. (a) : Cane Sugar, X, W1 = 5 g, W2 = 1 g, V1 = 100 mL, V2 = 100 mL, = 0.1 L, = 0.1 L, M1 = 342, M2 = ?, For isotonic solutions, C1 = C2, W2, W1, 5 1, M V  M V  342 0.1 M  0.1, 1 1, , 2 2, , M, , 2, , 342, ,  68.4, 5, 43. (b) : x2 (mole fraction of solute) = 0.2, From Raoult’s law,, 10, p ps,  x 2   0.2  p° = 50 mm Hg, p, p, Again, when p° – ps = 20 mm Hg, then, p ps = mole fraction of solute = 20 = 0.4, p, 50,  mole fraction of solvent = 1 – 0.4 = 0.6, 44. (c) : Vapour pressure of pure solvent (pA° ) =, 143 mm Hg, weight of solute (wB) = 0.5 g, weight, of solvent (WA) = 100 g, molecular weight of solute, (MB) = 65 and molecular weight of solvent (MA) = 154., 2, , p, A p, p, A, , or, , p, , s, , s, , w M, 143  ps  0.5 154,  MB WA or, 143  65 100, B A, , = 141.31 mm Hg, , www.neetujee.com, , 45. (a) : Weight of glucose = 10 g,, Weight of urea = 10 g and weight of sucrose = 10 g, The number of moles of glucose, Weight, 10, (n1) =,  = 0.05, Molecular weight 180, 10, Similarly, number of moles of urea (n2) =, = 0.16 and, 10 60, the number of moles of sucrose (n ) =, = 0.03, 3, 342, The osmotic pressure is a colligative property and it, depends upon the number of moles of a solute., Since n2 > n1 > n3, therefore p2 > p1 > p3., 46. (a), 47. (a) : There is no net movement of the solvent, through the semipermeable membrane between two, solutions of equal concentration., 48. (c) : The properties which depend only upon, the number of solute particles present in the solution, irrespective of their nature are called colligative, properties. Lowering in vapour pressure, elevation in, boiling point, depression in freezing point and osmotic, pressure are colligative properties., 49. (b) : Blood cells neither swell nor shrink in isotonic, solution. The solutions having same osmotic pressure are, called isotonic solutions., 50. (b) : Relative lowering of vapour pressure is equal, to mole fraction of solute which is the ratio of solute, molecules to the total molecules in solution., 51. (d) : Being a strong electrolyte, Ba(OH)2 undergoes, 100% dissociation in a 2+, dilute aqueous solution,, Ba(OH)2(aq)  Ba (aq) + 2OH– aq), (, Thus, van’t Hoff factor i = 3., 52. (c) : Tb = iKb m, For equimolal solutions, elevation in boiling point will, be higher if solution undergoes dissociation i.e., i > 1., 53. (c) : Tf = i × Kf × m, So, Tf  i (van’t Hoff factor), Salt, i, KCl, 2, C6H12O6, 1, Al2(SO4) 3, 5, K2SO4, 3, Hence, i is maximum i.e., 5 for Al2(SO4)3., 54. (c) : From the value of van’t Hoff factor i it is possible, to determine the degree of dissociation or association., In case of dissociation, i is greater than 1 and in case of, association i is less than 1., 55. (b) : We know that, wB 1000, Tf  i  K f , M W, B, , A, , www.mediit.in

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16, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , Given : Tf = 3.82, Kf = 1.86,, wB = 5, MB = 142, WA = 45, Tf  MB  WA 3.82 142  45, i, ,  2.63, K f  wB 1000 1.86  5 1000, , Total = 1 + ,  i = 1 +  = 1 + 0.2 = 1.2, Tf = i × Kf × m = 1.2 × 1.86 × 0.5 = 1.116 K  1.12 K, , 56. (d) : We know that Tf = i × Kf × m, Here i is van’t Hoff factor., i for weak acid is 1 + ., Here  is degree of dissociation i.e., 30/100 = 0.3,  i = 1 +  = 1 + 0.3 = 1.3, Tf = i × Kf × m = 1.3 × 1.86 × 0.1 = 0.24,  Freezing point of solution, Tf = T°f – Tf, = 0 – 0.24 = – 0.24°C, 57. (d) : Addition of water to an aqueous solution of, KI causes the concentration of the solution to decrease, thereby increasing the vapour pressure. In the other, three options, the electrolytes undergo ionization, which, leads to lowering of vapour pressure., 58. (d) : The number of moles of ions produced by, 1 mol of ionic compound = i, Applying, Tf = i × Kf × m, 0.00732 = i × 1.86 × 0.002, 0.00732, i,  1.96  2, 1.86  0.002, 59. (b) : HX  H+ + X –, 1–, , , , 60. (c) : Since Al2(SO4) 3 gives maximum number of, ions on dissociation, therefore it will have the lowest, freezing point., Tf = iKf · m, 61. (b) : K3[Fe(CN)6]  3K+ + [Fe(CN)6]3- and, –, Al(NO3)3  Al3+ + 3NO, 3, Since both Al(NO3)3 and K3[Fe(CN)6] give the same, number of ions, therefore they have the same van’t Hoff, factor., 62. (c) : In solution, CaCl2 gives three ions, KCl gives, two ions while glucose and urea are covalent molecules so, they do not undergo ionisation. Since osmotic pressure is, a colligative property and it depends upon the number of, solute particles (ions), therefore, 0.1 M solution of CaCl2, exhibits the highest osmotic pressure., 63. (a) : Here, Tf = i × Kf × m, van’t Hoff factor, i = 2 for NaCl, so conc. = 0.02, which is, maximum in the present case., Hence, Tf is maximum or freezing point is minimum, in 0.01 m NaCl., , , , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 3, , 3.2, 1., , 2., , Electrochemistry, standard state are connected to make a cell. The cell, potential will be, (a) + 1.19 V, (b) + 0.89 V, (c) + 0.18 V, (d) + 1.83 V, (2011), , Galvanic Cells, , The standard electrode potential (E°) values of, Al3+/Al, Ag+/Ag, K+/K and Cr3+/Cr are –1.66 V,, 0.80 V, –2.93 V and –0.74 V, respectively. The correct, decreasing order of reducing power of the metal is, (a) Ag > Cr > Al > K, (b) K > Al > Cr > Ag, (c) K > Al > Ag > Cr, (d) Al > K > Ag > Cr, (Odisha NEET 2019), , 6., , A button cell used in watches function2+, as following :, Zn(s) + Ag2O(s) + H2O(l)  2Ag(s) + Zn (aq) + 2OH–, (aq), , If half cell potentials, are, –, Zn2+ (aq) + 2e  Zn (s); E° = – 0.76 V, , 7., , Ag2O(s) + H2O(l) + 2e–  2Ag(s) + 2OH–(aq); E° = 0.34 V, The cell potential will be, (a) 0.84 V, (b) 1.34 V, (c) 1.10 V, (d) 0.42 V (NEET 2013), 3., , Cl2(g) + 2e–  2Cl–(aq) ; E° = + 1.36 V, Br2(l) + 2e–  2Br–(aq) ; E° = + 1.06 V, , 4., , 5., , Standard electrode potentials of three metals X, Y, and Z are –1.2 V, + 0.5 V and – 3.0 V respectively., The reducing power of these metals will be, (a) Y > Z > X, (b) Y > X > Z, (c) Z > X > Y, (d) X > Y > Z, (2011), Standard electrode potential for Sn4+/Sn2+, couple is +0.15 V and that for the Cr 3+/Cr, couple is –0.74 V. These two couples in their, , www.neetujee.com, , (d) Fe2+ will be oxidised to Fe3+., (Mains 2011), Consider the following relations for emf of an, electrochemical cell, (i) EMF of cell = (Oxidation potential of anode) –, (Reduction potential of cathode), (ii) EMF of cell = (Oxidation potential of anode) +, (Reduction potential of cathode), (iii) EMF of cell = (Reductional potential of anode), + (Reduction potential of cathode), (iv) EMF of cell = (Oxidation potential of anode) –, (Oxidation potential of cathode), Which of the above relations are correct?, (a) (iii) and (i), (b) (i) and (ii), (c) (iii) and (iv), (d) (ii) and (iv), (Mains 2010), , Standard reduction potentials of the half reactions, are given below :, F2(g) + 2e–  2F– (aq) ; E° = + 2.85 V, , I2(s) + 2e–  2I– aq), ; E° = + 0.53 V, (, The strongest oxidising and reducing agents, respectively are, (a) F2 and I–, (b) Br 2 and Cl–, (c) Cl2 and Br–, (d) Cl2 and I2, (Mains 2012), , A solution contains Fe2+, Fe3+ and I– ions. This, solution was treated with iodine at 35°C. E° for, Fe3+/Fe2+ is + 0.77 V and E° for I2/2I– = 0.536 V. The, favourable redox reaction is, (a) I2 will be reduced to I–, (b) there, will be no redox reaction, (c) I– will be oxidised to I2, , 8., , 9., , On the basis of the following E° values, the strongest, oxidizing agent is, [Fe(CN)6]4–  [Fe(CN)6]3– + e– ; E° = –0.35 V, Fe2+  Fe3+ + e– ; E° = –0.77 V, (a) Fe3+, (b) [Fe(CN)6]3–, (c) [Fe(CN)6]4–, (d) Fe2+, (2008), A hypothetical electrochemical cell is shown below :, A / A (x M)|| B( y M)| B, The emf measured is + 0.20 V. The cell reaction is, (a) A + B+  A+ + B, (b) A+ + B  A + B+, , www.mediit.in

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18, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (c) A+ + e–  A; B+ + e–  B, (d) the cell reaction cannot be predicted., , (2006), , 10. E°Fe2+/Fe = – 0.441 V and E°Fe3+/Fe2+ = 0.771 V, the, standard EMF of the reaction Fe + 2Fe3+  3Fe2+, will be, (a) 0.111 V, (b) 0.330 V, (c) 1.653 V, (d) 1.212 V, (2006), 11. Standard electrode potentials are, Fe2+/Fe;, 3+, 2+, 2+, 3+, E° = –0.44 and Fe /Fe ; E° = 0.77. Fe , Fe and Fe, blocks are kept together, then, (a) Fe3+ increases, (b) Fe3+ decreases, 2+, 3+, (c) Fe /Fe remains unchanged, (d) Fe2+ decreases., (2001), 12. Electrode potential for the following half-cell, reactions are, Zn  Zn2+ + 2e–; E° = + 0.76 V;, Fe  Fe2+ + 2e–; E° = + 0.44 V., The EMF for the cell reaction, Fe2+ + Zn  Zn2+ + Fe will be, (a) – 0.32 V, (b) + 1.20 V, (c) – 1.20 V, (d) + 0.32 V, (1996), 13. An electrochemical cell is set up as :, Pt; H2 (1 atm)|HCl(0.1 M) || CH3COOH (0.1 M), |H2 (1 atm); Pt. The e.m.f. of this cell will not be, zero, because, (a) acids used in two compartments are different, (b) e.m.f. depends on molarities of acids used, (c) the temperature is constant, (d) pH of 0.1 M HCl and 0.1 M CH COOH is not, 3, same., (1995), 14. Standard reduction potentials at 25°C of Li+|Li,, Ba2+|Ba, Na+|Na and Mg2+|Mg are –3.05, –2.90,, –2.71 and –2.37 volt respectively. Which one of the, following is the strongest oxidising agent?, (a) Ba2+, (b) Mg2+, +, (c) Na, (d) Li+, (1994), 15. A solution of potassium bromide is treated with, each of the following. Which one would liberate, bromine?, (a) Hydrogen iodide, (b) Sulphur dioxide, (c) Chlorine, (d) Iodine, (1993), , 3.3, , Nernst Equation, , 16. For the cell reaction :, 2Fe3+ + 2I–, 2Fe2+ + I, (aq), , (aq), , (aq), , 2(aq), , E°cell = 0.24 V at 298 K. The standard Gibbs’ energy, (rG°)of the cell reaction is, [Given that Faraday constant, F = 96500 C mol–1], , www.neetujee.com, , (a) 23.16 kJ mol–1, (c) –23.16 kJ mol–1, , (b) –46.32 kJ mol–1, (d) 46.32 kJ mol–1, (NEET 2019), , 17. For a cell involving one electron, E°cell = 0.59 V at, 298 K, the equilibrium constant for the cell reaction, 2.303RT, is [Given that,  0.059 V at T = 298 K], F, (b) 1.0 × 102, (a) 1.0 × 1030, 5, (c) 1.0 × 10, (d) 1.0 × 1010, (NEET 2019), 18. In the electrochemical cell :, Zn|ZnSO4(0.01 M)||CuSO4(1.0 M)|Cu,, the emf of this Daniell cell is E1. When the, concentration of ZnSO4 is changed to 1.0 M and that, of CuSO4 changed to 0.01 M, the emf changes to E2., From the followings, which one is the relationship, between E1 and E2? (Given, RT/F = 0.059), (a) E1 < E2, (b) E1 > E2, (d) E1 = E2, (c) E2 = 0  E1, (NEET 2017, 2003), 19. If the E°cell for a given reaction has a negative value,, which of the following gives the correct relationships, for the values of G° and Keq ?, (a) G° > 0; Keq < 1 (b) G° > 0; Keq > 1, (c) G° < 0; Keq > 1, , (d) G° < 0; Keq < 1, (NEET-II 2016, 2011), , 20. The pressure of H2 required to make the potential of, H2 electrode zero in pure water at 298 K is, (a) 10–10 atm, (b) 10–4 atm, (c) 10–14 atm, (d) 10–12 atm., (NEET-I 2016), 21. A hydrogen gas electrode is made by dipping, platinum wire in a solution of HCl of pH = 10 and, by passing hydrogen gas around the platinum wire, at one atm pressure. The oxidation potential of, electrode would be, (a) 0.118 V, (b) 1.18 V, (c) 0.059 V, (d) 0.59 V (NEET 2013), 22. Consider the half-cell reduction reaction, Mn2+ + 2e–  Mn, E° = – 1.18 V, Mn2+  Mn3+ + e– , E° = – 1.51 V, The E° for the reaction,, 3Mn2+  Mn0 + 2Mn3+,, and possibility of the forward reaction are, respectively, (a) – 4.18 V and yes, (b) + 0.33 V and yes, (c) + 2.69 V and no, (d) – 2.69 V and no., (Karnataka NEET 2013), 23. The Gibbs’ energy for the decomposition of Al2O3 at, 500 °C is as follows, , www.mediit.in

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Electrochemistry, 2, 4, Al2O3  Al  O2, rG = + 960 kJ mol–1, , 19, , 3, 3, The potential difference needed for the electrolytic, reduction of aluminium oxide (Al 2O3) at 500 °C is, at least, (a) 4.5 V, (b) 3.0 V, (c) 2.5 V, (d) 5.0 V (Mains 2012), 24. The electrode, potentials, for,, –, +, and Cu+ + e–  Cu, Cu2+ + e  Cu, (aq), , (aq), , (aq), , (s), , are + 0.15 V and + 0.50 V respectively. The value of, E°Cu2+/Cu will be, (a) 0.500 V, (b) 0.325 V, (c) 0.650 V, (d) 0.150 V, (2011), 25. For the reduction of silver ions with copper metal,, the standard cell potential was found to be + 0.46 V, at 25 °C. The value of standard Gibbs energy, G°, will be (F = 96500 C mol–1), (a) – 89.0 kJ, (b) – 89.0 J, (c) – 44.5 kJ, (d) – 98.0 kJ, (2010), 26. Given :, (i) Cu2+ + 2e–  Cu, E° = 0.337 V, (ii) Cu2+ + e–  Cu+, E° = 0.153 V, Electrode potential, E° for the reaction,, Cu+ + e–  Cu, will be, (a) 0.90 V, (b) 0.30 V, (c) 0.38 V, (d) 0.52 V, (2009), 27. Standard free energies of formation (in kJ/mol) at, 298 K are –237.2, –394.4 and –8.2 for H O , CO, 2(g), and pentane respectively. The value of2 (l), E°, for, the, (g), cell, pentane-oxygen fuel cell is, (a) 1.0968 V, (b) 0.0968 V, (c) 1.968 V, (d) 2.0968 V, (2008), 28. The equilibrium constant of the reaction :, Cu + 2Ag+,  Cu2+ (aq) + 2Ag(s);, (aq), E (s), ° = 0.46 V at 298 K is, 10, (a) 2.0 × 1010, (b) 4.0 × 10, (c) 4.0 × 1015, (d) 2.4 × 1010, (2007), 29. The standard e.m.f. of a galvanic cell involving cell, reaction with n = 2 is found to be 0.295 V at 25°C., The equilibrium constant of the reaction would be, (a) 2.0 × 1011, (b) 4.0 × 1012, 2, (c) 1.0 × 10, (d) 1.0 × 1010, –1, (Given F = 96500 C mol , R = 8.314 J K–1 mol–1), (2004), 30. On the basis of the information available from the, reaction,, 4/3Al + O2  2/3Al2O3, G = –827 kJ mol–1, of O2, the minimum e.m.f. required to carry out an, electrolysis of Al2O3 is (F = 96500 C mol–1), , www.neetujee.com, , (a) 2.14 V, (b) 4.28 V, (c) 6.42 V, (d) 8.56 V, (2003), 31. For, the, disproportionation, of, copper, 2Cu+  Cu2+ + Cu, E° is (Given : E° for Cu2+/Cu is, 0.34 V and E° for Cu2+/Cu+ is 0.15 V), (a) 0.49 V, (b) –0.19 V, (c) 0.38 V, (d) –0.38 V, (2000), 32. E° for the cell, Zn | Zn2+ ||Cu2+ | Cu is 1.10 V, (aq), , (aq), , at 25°C, the equilibrium constant for the reaction, is of the order, Zn + Cu2+(aq)  Cu + Zn2+, (aq), (a) 10+18, (b) 10+17, (c) 10–28, (d) 10+37, (1997), , 3.4, , Conductance of Electrolytic Solutions, , 33. Following limiting molar conductivities are given, as :, m(H2SO, = x S cm2 mol–1, 4 ), m(K 2SO4 ) = y S cm2 mol–1, m(CH3COOK) = z S cm2 mol –1, , m (in S cm2 mol–1) for CH3COOH will be, , (a) x – y + 2z, (c) x – y + z, , (b) x + y – z, (x  y), z, (d), 2, (Odisha NEET 2019), , 34. The molar conductivity of a 0.5 mol/dm3 solution, of AgNO3 with electrolytic conductivity of, –1, 5.76 × 10–3 S cm, at 298 K is, (a) 2.88 S cm2/mol, (b) 11.52 S cm2/mol, (c) 0.086 S cm2/mol, , (d) 28.8 S cm2/mol, (NEET-II 2016), , 35. At 25 °C molar conductance of 0.1 molar, aqueous solution of ammonium hydroxide is, 9.54 ohm–1 cm2 mol–1 and at infinite, dilution its, molar conductance is 238 ohm–1 cm2 mol–1. The, degree of ionisation of ammonium hydroxide at the, same concentration and temperature is, (a) 4.008%, (b) 40.800%, (c) 2.080%, (d) 20.800% (NEET 2013), 36. Limiting molar conductivity of NH4OH, [i.e., °, ] is equal to, m(NH4OH), , (a), (b), (c), (d), , °m(NH4Cl) + °m(NaCl) – °m(NaOH), °m(NaOH) + °m(NaCl) – m° (NH4Cl), °m(NH4OH) + °m(NH4Cl) – °m(HCl), °m(NH4Cl) + °m(NaOH) – °m(NaCl), , (2012), , 37. Molar conductivities (°m) at infinite dilution of, NaCl, HCl and CH3COONa are 126.4, 425.9 and, 91.0 S cm2 mol–1 respectively. (°m) for CH3COOH, will be, , www.mediit.in

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20, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, –1, , 2, , (a) 425.5 S cm mol, (c) 290.8 S cm2 mol–1, , 2, , –1, , (b) 180.5 S cm mol, (d) 390.5 S cm2 mol–1, (Mains 2012, 1997), , 38. An increase in equivalent conductance of a strong, electrolyte with dilution is mainly due to, (a) increase in ionic mobility of ions, (b) 100% ionisation of electrolyte at normal dilution, (c) increase in both i.e., number of ions and ionic, mobility of ions, (d) increase in number of ions., (2010), 39. Which of the following expressions correctly, represents the equivalent conductance at infinite, dilution of Al (SO ) ? Given that  3 and  2, 2, , 4 3, , Al, , SO, 4, , are the equivalent conductances at infinite dilution, of the respective ions., (a) 2 3  3 2, Al, SO, 4, ,  2, (b) Al3  , SO, 4, , (c) (  3   2 )  6, Al, SO 4, 1, 1, (d) , , 3  SO2, 2, 3 Al, 4, , 44. On heating one end of a piece of a metal, the other, end becomes hot because of, (a) energised electrons moving to the other end, (b) minor perturbation in the energy of atoms, (c) resistance of the metal, (d) mobility of atoms in the metal., (1995), , 3.5, , Electrolytic Cells and Electrolysis, , 45. On electrolysis of dil. sulphuric acid using platinum, (Pt) electrode, the product obtained at anode will be, (a) hydrogen gas, (b) oxygen gas, (c) H2S gas, (d) SO2 gas. (NEET 2020), 46. The number of Faradays (F) required to produce, 20 g of calcium from molten CaCl (Atomic mass of, 2, , (Mains 2010), , 40. The equivalent conductance of M/32 solution of a, weak monobasic acid is 8.0 mho cm2 and at infinite, dilution is 400 mho cm2. The dissociation constant, of this acid is, (a) 1.25 × 10–6, (b) 6.25 × 10–4, (c) 1.25 × 10–4, (d) 1.25 × 10–5, (2009), 41. Kohlrausch’s law states that at, (a) infinite dilution, each ion makes definite, contribution to conductance of an electrolyte, whatever be the nature of the other ion of the, electrolyte, (b) infinite dilution, each ion makes definite, contribution to equivalent conductance of an, electrolyte, whatever be the nature of the other, ion of the electrolyte, (c) finite dilution, each ion makes definite, contribution to equivalent conductance of an, electrolyte, whatever be the nature of the other, ion of the electrolyte, (d) infinite dilution each ion makes definite, contribution to equivalent conductance of an, electrolyte depending on the nature of the other, ion of the electrolyte., (2008), 42. Equivalent conductances of Ba2+ and Cl– ions are, 127 and 76 ohm–1 cm–1 eq–1 respectively. Equivalent, conductance of BaCl2 at infinite dilution is, (a) 139.5, (b) 101.5, (c) 203, (d) 279, (2000), , www.neetujee.com, , 43. The specific conductance of a 0.1 N KCl solution, at 23°C is 0.012 ohm–1 cm–1. The resistance of cell, containing the solution at the same temperature was, found to be 55 ohm. The cell constant will be, (a) 0.918 cm–1, (b) 0.66 cm–1, –1, (c) 1.142 cm, (d) 1.12 cm–1, (1999), , Ca = 40 g mol–1) is, (a) 1, (b) 2, (c) 3, (d) 4, (NEET 2020), 47. During the electrolysis of molten sodium chloride,, the time required to produce 0.10 mol of chlorine, gas using a current of 3 amperes is, (a) 55 minutes, (b) 110 minutes, (c) 220 minutes, (d) 330 minutes., (NEET-II 2016), 48. The number of electrons delivered at the cathode, during electrolysis by a current of 1 ampere in, 60 seconds is (charge on electron = 1.60 × 10–19 C), (a) 6 × 1023, (b) 6 × 1020, 20, (c) 3.75 × 10, (d) 7.48 × 1023, (NEET-II 2016), 49. When 0.1 mol MnO2–, 4 is oxidised, the quantity of, 2–, , electricity required to completely oxidise MnO4 to, MnO–4 is, (a) 96500 C, (b) 2 × 96500 C, (c) 9650 C, (d) 96.50 C, (2014), 50. The weight of silver (at. wt. = 108) displaced by a, quantity of electricity which displaces 5600 mL of, O2 at STP will be, (a) 5.4 g, (b) 10.8 g, (c) 54.0 g, (d) 108.0 g, (2014), 51. How many grams of cobalt metal will be deposited, when a solution of cobalt(II) chloride is electrolyzed, with a current of 10 amperes for 109 minutes?, (1 Faraday = 96,500 C; Atomic mass of Co = 59 u), (a) 4.0, (b) 20.0, (c) 40.0, (d) 0.66, (Karnataka NEET 2013), , www.mediit.in

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Electrochemistry, , 21, , 52. Al2O3 is reduced by electrolysis at low potentials, and high currents. If 4.0 × 104 amperes of current, is passed through molten Al 2O3 for 6 hours, what, mass of aluminium is produced? (Assume 100%, current efficiency, at. mass of Al = 27 g mol–1), (a) 8.1 × 104 g, (b) 2.4 × 105 g, (c) 1.3 × 104 g, (d) 9.0 × 103 g (2009), 53. 4.5 g of aluminium (at. mass 27 amu) is deposited at, cathode from Al3+ solution by a certain quantity of, electric charge. The volume of hydrogen produced at, STP from H + ions in solution by the same quantity, of electric charge will be, (a) 44.8 L, (b) 22.4 L, (c) 11.2 L, (d) 5.6 L, (2005), 54. In electrolysis of NaCl when Pt electrode is taken, then H2 is liberated at cathode while with Hg, cathode it forms sodium amalgam. The reason for, this is, (a) Hg is more inert than Pt, (b) more voltage is required to reduce H+ at Hg, than at Pt, (c) Na is dissolved in Hg while it does not dissolve, in Pt, (d) conc. of H+ ions is larger when Pt electrode is, taken., (2002), 55. A 5 ampere current is passed through a solution of, zinc sulphate for 40 minutes. The amount of zinc, deposited at the cathode is, (a) 0.4065 g, (b) 65.04 g, (c) 40.65 g, (d) 4.065 g, (1996), 56. Sodium is made by the electrolysis of a molten, mixture of about 40% NaCl and 60% CaCl2 because, (a) Ca++ can reduce NaCl to Na, (b) Ca++ can displace Na from NaCl, (c) CaCl2 helps in conduction of electricity, (d) this mixture has a lower melting point than, NaCl., (1995), 57. When CuSO4 is electrolysed using platinum, electrodes,, , (a), (b), (c), (d), , copper is liberated at cathode, sulphur at anode, copper is liberated at cathode, oxygen at anode, sulphur is liberated at cathode, oxygen at anode, oxygen is liberated at cathode, copper at anode., (1993), 58. On electrolysis of dilute sulphuric acid using, platinum electrodes, the product obtained at the, anode will be, (a) hydrogen, (b) oxygen, (c) hydrogen sulphide (d) sulphur dioxide., (1992), , 3.7, , Fuel Cells, , 59. A device that converts energy of combustion of fuels, like hydrogen and methane, directly into electrical, energy is known as, (a) dynamo, (b) Ni-Cd cell, (c) fuel cell, (d) electrolytic cell., (2015, Cancelled), 60. The efficiency of a fuel cell is given by, (a) G/S, (b) G/H, (c) S/G, (d) H/G, , 3.8, , (2007), , Corrosion, , 61. Zinc can be coated on iron to produce galvanized, iron but the reverse is not possible. It is because, (a) zinc is lighter than iron, (b) zinc has lower melting point than iron, (c) zinc has lower negative electrode potential than, iron, (d) zinc has higher negative electrode potential, than iron., (NEET-II 2016), 62. The most convenient method to protect the bottom, of ship made of iron is, (a) coating it with red lead oxide, (b) white tin plating, (c) connecting it with Mg block, (d) connecting it with Pb block., (2001), 63. To protect iron against corrosion, the most durable, metal plating on it, is, (a) copper plating, (b) zinc plating, (c) nickel plating, (d) tin plating., (1994), , ANSWER KEY, , 1., 11., , (b), , (c), , (b), , 2., 12., , (a), , (d), , 3., 13., , (c), , (d), , 4., 14., , 21., , (d), , 22., , (d), , 23., , (c), , 31., , (c), , 32., , (d), , 33., , 41., , (a), , 42., , (a), , 51., , (b), , 52., , 61., , (d), , 62., , www.neetujee.com, , (b), , (b), , 5., 15., , (d), , (b), , 7., 17., , 24., , (b), , (d), , 34., , 43., , (b), , (a), , 53., , (d), , (b), , 63., , (b), , (a), , (d), , 8., 18., , 26., , (d), , 27., , (a), , (a), , 36., , (d), , 37., , 45., , (b), , 46., , (a), , 55., , (d), , 56., , (d), , (c), , 6., 16., , 25., , (a), , (b), , 35., , 44., , (a), , 54., , (b), , (c), , (a), , (b), , 9., 19., , (d), , (a), , 10., 20., , 28., , (c), , 29., , (d), , 30., , (a), , (d), , 38., , (a), , 39., , (b), , 40., , (d), , 47., , (b), , 48., , (c), , 49., , (c), , 50., , (d), , 57., , (b), , 58., , (b), , 59., , (c), , 60., , (b), , (c), , www.mediit.in

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22, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , Hints & Explanations, 1. (b) : Higher the value of E°red, stronger is the, oxidising power. Thus, the decreasing order of reducing, power of the metal is K > Al > Cr > Ag., 2. (c) : E°cell = E°O.P. + E°R.P., = 0.76 + 0.34 = 1.10 V, 3. (a) : Less positive the value of reducti on potential,, –, stronger will be the reducing agent thus, I is strongest, reducing agent. More positive, the value of reduction, potential shows good oxidising properties thus, strongest, oxidising agent is F2., , If a cell is constructed combining these two electrodes, oxidation occurs at Fe2+/Fe electrode., At anode : Fe  Fe2+ + 2e–, At cathode : [Fe3+ + e–  Fe2+] × 2, , 4. (c) : More negative the value of reduction potential,, stronger will be the reducing agent., So, Z (–3.0 V) > X (–1.2 V) > Y (+ 0.5 V), 5. (b) : E°cell = E°cathode – E°anode, = 0.15 – (– 0.74) = 0.15 + 0.74 = 0.89 V, , 12. (d) : E°Zn/Zn2+ = +0.76 V, E°Fe/Fe2+ = 0.44 V  E°Fe2+/Fe = –0.44 V, E°cell = E°O.P. + E°R.P. = + 0.76 – 0.44 = + 0.32 V, 13. (d) : Since it is a concentration cell and the, concentration of H+ ions in two electrolyte solutions, (HCl and CH3COOH) are different i.e., pH of 0.1 M HCl, and 0.1 M CH3COOH is not same, therefore e.m.f. of this, cell will not be zero., , 6. (c) : Since the reduction, potential, of, Fe3+/Fe2+ is greater than that of I2/I–, Fe3+ will be reduced, and I– will be oxidised., 2Fe3+ + 2I– 2Fe2+ + I2, 7. (d) : EMF of a cell = Reduction potential of cathode, – Reduction potential of anode, = Reduction potential of cathode +, Oxidation potential of anode, = Oxidation potential of anode –, Oxidation potential of cathode., (a) : [Fe(CN)6]3–  [Fe(CN)6]4–, E° = +0.35 V, Fe3+  Fe2+; E° = +0.77 V, Higher the +ve reduction potential, stronger will be, the oxidising agent. Oxidising agent oxidises other, compounds and gets itself reduced easily. Thus, Fe3+ is, the strongest oxidising agent., 9. (a) : From the given expression :, At anode : A, A+ + e– (oxidation), +, At cathode : B + e–, B (reduction), Overall reaction is : A + B+, A+ + B, 2+, –, 10. (d) : Fe + 2e, Fe ; E° = –0.441 V, ... (i), 8., , Fe3+ + e–, Fe2+ ; E° = 0.771 V, 3+, Fe + 2Fe, 3Fe2+ ; E° = ?, To get the above equation, (ii) × 2 – (i), 2Fe3+ + 2e–, 2Fe2+ ; E° = 0.771 V, 2+, –, –Fe – 2e, – Fe ; E° = +0.441 V, 2Fe3+ + Fe, , 3Fe2+ ; E° = 1.212 V, , 11. (b) : E 2,   0.44 V, Fe /Fe, E 3+ 2   0.77 V, Fe /Fe, , www.neetujee.com, , ... (ii), , Cell reaction : Fe + 2Fe3+  3Fe2+, If Fe2+, Fe3+ and Fe blocks are kept together then Fe3+, reacts with Fe to yield Fe2+ i.e., concentration of Fe3+ is, decreased and that of Fe2+ is increased., , 14. (b) : More positive or less negative the reduction, potential value, the stronger is the oxidising agent., 15. (c) : A stronger oxidising agent (Cl 2) displaces a, weaker oxidising agent (Br2) from its salt solution., 2KBr + Cl2  2KCl + Br2, 16. (b) : The standard Gibbs’ energy,, (G°) = –nFE°cell. Value of n = 2,  G° = – 2 × 96500 × 0.24 = – 46320 J, = – 46.32 kJ/mol, 17. (d) : According to Nernst equation,, 0.059, , Ecell  Ecell , log Qc, n, At equilibrium Ecell = 0,  Qc = Kc, 0.059, 0.059, , Ecell  n log Kc  0.59  1 log Kc, K = antilog 10  K = 1 × 1010, c, , c, , 18. (b) : Ecell = E°cell –, , 0.059, , log, , [Zn 2, ], , , , n, [Cu2 ], 0.059, 0.01, E1  E , log, 2, 1, 0.059, E = E° –, (–2) = E° + 0.059, 1, 2, 0.059, 1, E2  E , log  E 0.059, 0.01, 2, Hence, E1 > E2., 19. (a) : G° = –nFE°cell, If E°cell = – ve then G° = +ve i.e.; G° > 0., , www.mediit.in

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Electrochemistry, , 23, , G° = –nRT log Keq, For G° = +ve, Keq = –ve i.e., Keq < 1., , 25. (a) : The cell reaction can be written as, Cu + 2Ag+, Cu2+ + 2Ag, We know, G° = – nFE°cell, = – 2 × 96500 × 0.46 = – 88780 J, = – 88.78 kJ  – 89 kJ, 26. (d) : Given,, Cu ; E1° = 0.337 V, Cu2+ + 2e–, 2+, –, Cu + e, Cu+ ; E°2 = 0.153 V, The required reaction is, Cu+ + e–, Cu ;, E3° = ?, , 20. (c) : pH = 7 for water., –log[H+] = 7  [H+] = 10–7, 2H+ (aq) + 2e–, H2(g), 0.0591, E  E , log pH2, cell, cell, 2, [H+ ]2, H2, 0.0591 log p, 2, (10 7 )2, pH2, pH2, 0 , 1, log, (107 )2, (107 )2, pH = 10–14 atm, , 00, , [Q log 1 = 0], , 3, , or, , 10, 0.059, log, 1, 2, , 10, , EH, , 2H, , E, ,  0 , , H2 H+, , , , 23. (c) : G° = – nFE°, F = 96500, G° = + 960 × 103 J/mol, 2, , 4, Al2O3  Al  O2, 3, 3, Total number of Al atoms in Al2O3  2  2  4, 3, 3, Al3+ + 3e–, Al, As 3e– change occur for each Al-atom, 4, n, ,  3 4,  total, 3, G, 960 1000, E , , nF, 4  96500,  E° = –2.48  –2.5 V, 24. (b) : Cu2+, + e–, Cu+, ; E ° = 0.15 V, –, , (aq), , 1, , Cu (aq) + e, Cu(s) ; E2° = 0.50 V, Cu2+ + 2e–, Cu ; E° = ?, Now, G° = G1° + G2°, or, – nFE° = –n1FE1° – n2FE2°, n E  n E 1  0.15 1  0.50, or,, = 0.325 V, E 1 1 2 2 , n, 2, www.neetujee.com, , 2, , E3° = 2 × E1°– E2°, , 27. (a) : C5H12(g) + 8O2(g), , 22. (d) : Mn2+ + 2e–, Mn; E° = –1.18 V, ... (i), 2+, 3+, –, 2Mn, 2Mn + 2e ; E° = –1.51 V, ... (ii), By adding equation (i) and (ii), we get equation for the, cell,, 3Mn2+, Mn + 2Mn3+; E° = –2.69 V, Since the E° value is negative, so the process is nonspontaneous as G° is positive., , (aq), , 2, , 5CO2(g) + 6H2O(l), , G° = [(–394.4 × 5) + (–237.2 × 6)] – [(–8.2) + (8 × 0)], , 2, ,  0.59 V, , +, , 1, , or E  (2  0.337)  0.153  0.52 V, , 21. (d) : H2 ,  2H  2e, 1010, , 1, , 3, , 2, , 1 atm, , Applying, G° = –nFE°, G3  G1  G2, (n3FE )  (n FE)  (n FE), , = –3387 kJ, Note that the standard free energy change of elementary, substances is taken as zero., For the fuel cell, the complete cell reaction is :, C5H12(g) + 8O2(g), 5CO2(g) + 6H2O(l), which is the combination of the following two half, reactions :, C5H12(g) + 10H2O(l), , 5CO2(g) + 32H+ + 32e– and, , 8O2(g) + 32H+ + 32e–, 16H2O(l), Therefore, the number of electrons exchanged is 32 here,, i.e., n = 32., G° = –nFE° = – 3387 × 103 J, = –32 × 96500 J/Volt × E°, Thus, E° = 1.0968 V, 28. (c) : For a cell reaction in equilibrium at 298 K,, 0.0591, E , log K, cell, c, n, where, Kc = equilibrium constant, n = number of, electrons involved in the electrochemical cell reaction., Given, Ec°ell = 0.46 V, n = 2, 2  0.46, 0.0591,  15.57,  0.46 ,  log Kc or, log Kc , 0.0591, 2, or, Kc = 3.7 × 1015  4 × 1015, 0.0591, 29. (d) : E  E , log10 Q at 25°C, n, At equilibrium, E = 0, Q = K, 0.0591, 0  E , log10 K, nnE, , , or, K  antilog, ,  0.0591, , www.mediit.in

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24, , or, K  antilog, ,  2  0.295,  0.0591 , ,  antilog, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry,  1000, 34. (b) : m , Molarity( M), ,  0.590 , ,  0.0591, , = antilog 10 = 1 × 1010, , , , 30. (a) : G° = –nFE°, G 827000, E nF  4  96500,  2.14 V, 4, ,  4, ,  , , 1, Al, , 3e, ,, Al, , , 3e, , 4e, , , , 3, 3, , , Cu2+ + e–, , Cu+; E2  0 15 V, , ...(2), , Cu+ + e–, , Cu; E3° = ?, , ...(3), , Now G1   nFE1  2  0  34  F = –0.68 V, G2  1  0 15  F, G  1  E  F ,, , = 0.53 – 0.15 = 0.38 V, 32. (d) : Nernst equation is, 0.059, E  E , log K, 2, 0.059,  E , log K (E = 0 at equilibrium condition), 2, 0.059,  1.1 , log K  K = 1.9 × 10+37, 2, 33. (d) : According to Kohlrausch’s law,, o,  o for CH COOH =  o,    2, 3, , CH3COO, , H, ,  for H2SO4 = 2    2  x S cm mol, 2, , H, , 1, , ...(i), , SO4, ,  for K2SO4 = 2K  SO2  y S cm 2mol 1,  for CH3COOK = , , ...(ii), , 4, ,     z S cm2mol1, K, ...(iii), On adding equation (i) and 2 × (iii) and subtracting (ii),, we get, 2    2  2,  2  2  , , , , CH3COO, , H, , SO , , 2, , , , , , H, , , , K, , CH COO, , 4, , 3, ,  2CH, , ,  x  2z  y, 3COO , , H  CH COO , 3, , www.neetujee.com, , (x  y), 2, , HCl, , NaCl, , = 390.5 S cm2 mol–1, 38. (a) : Strong electrolytes are completely ionised at all, concentrations. On increasing dilution, the no. of ions, remains the same but the ionic mobility increases and, the equivalent conductance increases., , , , 2, , CH3COONa, , = 91.0 + 425.9 – 126.4, , anode (Cu /Cu ), , cathode(Cu /Cu), , m, , 37. (d) : °NaCl = 126.4 S cm2 mol–1, °HCl = 425.9 S cm2 mol2 –1 –1, °CH COONa = 91.0 S cm mol, + ° – °, ° 3, = °, CH3COOH, ,  – 0.68 F = –0.15 F – E°3 × F,  E3 0  68  0 15  0  53V, As, E  E,  E , +, , = 11.52 S cm2 mol–1, , Molar conductivity at infinite dilution (m ), 9.54 1 cm2 mol1, , = 0.04008 = 4.008%, 238 1 cm2 mol1, 36. (d), , 3, , Again G1  G2  G3, , cell, , 0.5 mol cm3, , 35. (a) : Degree of dissociation, Molar conductivity at conc. C(c ), m, () , , , , 31. (c) : For the reaction, 2Cu+, Cu2+ + Cu the, cathode is Cu+/Cu and anode is Cu+/Cu2+., Given, Cu2+ + 2e–, Cu; E1°= 0.34 V, ...(1), , 3, , 5.76 103 Scm1 1000, , z, , K+, , SO2, 4, , = x + 2z – y, , 39. (b) : At infinite dilution, when dissociation is, complete, each ion makes a definite contribution towards, molar conductance of the electrolyte irrespective of the, nature of the other ion with which it is associated., Hence, Al (SO )   3   2, 2, , Al , , 4 3, , SO4, , 40. (d) : Given,  = 8 mho cm2,  = 400 mho cm2, , Degree of dissociation, , , 8,  ,  2  102, , 400, Dissociation constant, K = C, Given, C = M/32, 1, 2, 2, –5,  K   2  10  2  10 = 1.25 × 10, 32, 41. (a) : At infinite dilution, when dissociation is, complete each ion makes a definite contribution towards, molar conductance of the electrolyte irrespective of the, nature of the other ion with which it is associated and, that the molar conductance of any electrolyte at infinite, dilution is given by the sum of the contributions of two, ions. This is called Kohlrausch’s law.,      ,, m, , , , , , where,  and , , , , , are molar ionic conductance at, , infinite dilution for cation and anion, respectively., , www.mediit.in

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25, , Electrochemistry, 42. (a) : , , 1  1 ,      , n1, n,   1, , So,  BaCl         , 2+, , , 2, Ba, 2, 1 Cl , 1,  127  76  63 5  76  139  5, 2, –1, , 44. (a) : Conductivity of heat in metals is due to the, presence of free electrons, which move due to increase in, temperature., 45. (b) : During electrolysis of dilute sulphuric acid the, following reaction takes place at anode., 2H2O(l)  O2(g) + 4H+ (aq) + 4e–; E°cell = +1.23 V, i.e., O2(g) will be liberated at anode., 46. (a) : Ca 2  2e , , Ca, 2F, 1F, , (s), (1 mole = 40 g), 20 g, , Thus, one Faraday is required to produce 20 g of calcium, from molten CaCl2., 47. (b) : During the electrolysis of molten sodium, chloride,, At cathode : 2Na+ + 2e–, 2Na, At anode : 2Cl–, Cl2 + 2e–, Net reaction : 2Na+ + 2Cl–, , 2Na + Cl2, , According to Faraday’s first law of electrolysis,, w=Z×I×t, E, w,  I t, 96500, No. of moles of Cl2 gas × Mol. wt. of Cl2 gas, , , Eq. wt. of Cl2 gas  I  t, , 96500, 35.5  3  t, 0.10  71  96500, 0.10  71  96500,  6433.33 sec, 35.5  3, 6433.33, t, min = 107.22 min  110 min, 60, 48. (c) : Q = I × t, Q = 1 × 60 = 60 C, t, , www.neetujee.com, , 6, , 7, , MnO2,  MnO 4  e, 4 , , 0.1 mol, , –1, , 43. (b) :  = 0.012 ohm cm, 1 1, R = 55 ohm  C   ohm1, R 55,  l  Specific Conductance, Cell Constant ,  a , Conductance, 0.012, , = 55 × 0.012 = 0.66 cm–1, 1/ 55, , (aq), , Now, 1.60 × 10–19 C  1 electron, 60,  60 C ,  3.75 1020 electrons, 1.6 1019, 49. (c) : The oxidation reaction is, , 0.1 mol, , Q = 0.1 × F = 0.1 × 96500 C = 9650 C, 50. (d) : According to Faraday’s second law,, 5600, W, W, W,  32, O2, Ag , Ag, or, ,  22400, E, 108, 8, E, WAgAg 8 O2, or, ,  W = 108 g, Ag, 108 8, ItE, 51. (b) : w , 96500, 10 109  60  59, ,  19.99  20 g, 96500  2, 52. (a) : Applying E = Z × 96500, 27,  Z  96500  Z  9, 3, 96500, Now applying the formula, w = Z × I × t, 9, w,  4 104  6  60  60 = 8.1 × 104 g, , 96500, 53. (d) : We know that,, 1 Faraday charge liberates 1 eq. of substance., This is the Faraday law., 27, eq. wt. of Al =  9, 3, wt. of Al 4.5, ,  0.5, No. of eq. of Al =, eq. wt., 9, No. of Faradays required = 0.5,  No. of eq. of H 2 produced = 0.5 eq., 22.4,  11.2 L, Volume occupied by 1 eq. of H2 =, 2,  Volume occupied by 0.5 eq. of H2 = 11.2 × 0.5, = 5.6 L at STP, 54. (b) : When sodium chloride is dissolved in water, it, ionises as NaCl, Na+ + Cl–., Water also dissociates as : H2O, H+ + OH–, During passing of electric current through this solution, using platinum electrode, Na+ and H+ ions move towards, cathode. However, only H+ ions are discharged more, readily than Na+ ions because of their low discharge, potential (in the electromotive series hydrogen is lower, than sodium). These H+ ions gain electrons and change, into neutral atoms., , www.mediit.in

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26, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , At cathode H + + e–, H, H + H, H2, Cl– and OH– ions move towards anode. Cl– ions lose, electrons and change into neutral atom., At anode, Cl– – e– Cl, Cl + Cl Cl2, If mercury is used as cathode, H+ ions are not discharged, at mercury cathode because mercury has a high hydrogen, overvoltage. Na+ ions are discharged at the cathode in, preference to H+ ions, yielding sodium, which dissolves, in mercury to form sodium amalgam., At cathode : Na+ + e– Na, 55. (d) : Current (I) = 5 ampere and, time (t) = 40 minutes = 2400 seconds., Amount of electricity passed (Q) = I × t, = 5 × 2400 = 12000 C, Now, Zn2+ + 2e–, Zn (1 mole = 65.39 g), Since, two charges (i.e., 2 × 96500 C) deposits 65.39 g of, zinc, therefore 12000 C will deposit, 65.39 12000, , = 4.065 g of zinc, 2  96500, 56. (d) : Sodium is obtained by electrolytic reduction of, its chloride. Melting point of chloride of sodium is high, so in order to lower its melting point, calcium chloride is, added to it., 57. (b) : CuSO4, H2O H+ + OH–, , Cu2+ + SO2–, 4, , At cathode : Cu2+ + 2e–  Cu, At anode : 4OH–  2H 2O + O 2 + 4e–, 58. (b) : During electrolysis of dilute sulphuric acid,, product obtained at anode will be oxygen., At anode : 4OH–, 2H2O + 2O 2 + 4e–, 59. (c), 60. (b) : The thermal efficiency,  of a fuel conversion, device is the amount of useful energy produced relative, to the change in enthalpy, H between the product and, feed streams., useful energy, , H, In an ideal case of an electrochemical converter, such, as a fuel cell, the change in Gibb’s free energy, G of, the reaction is available as useful electric energy at that, temperature of the conversion., G, Hence, , =, ideal, H, 61. (d) : Reduction potential values of E°Zn2+/Zn = –0.76 V, and E°Fe2+/Fe = –0.44 V, Thus, due to higher negative electrode potential value of, zinc than iron, iron cannot be coated on zinc., 62. (b) : The most convenient method to protect the, bottom of the ship made of iron is white tin plating, preventing the build up of barnacles., 63. (b), , , , www.neetujee.com, , www.mediit.in

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28, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (c) , , d[B], , (d), , dt, 8., , 3A, , 2B, rate of reaction, , (a) , (c) , , 9., , , , d[D], , (2006), , dt, d[B], , is equal to, dt, 2 d[A], (b) , , 3 d[A], , 3 dt, d[A], (d) 2, dt, , 2 dt, 1 d[A], , (2002), 3 dt, For the reaction,, H+ + BrO3– + 3Br–, 5Br2 + H2O, which of the following relations correctly represents, the consumption, of products?, d[Brformation, d[Br, ] 3 d[Br2 ], d[Br ], 3 and, 2], , , (b), (a), 5 dt, dt, 5 dt, dt, d[Br ], 5 d[Br2 ] (d) d[Br ] 5 d[Br2 ], , (c), dt, 3 dt, dt, , dt, 3, (2000), , 10. For the reaction H 2(g) + I2(g) 2HI(g), the rate of, reaction is expressed as,  H2  1   I2 ,  HI , (a), , , 2t, t, t,  I2 , (b) ,     H2   1 HI, 2 t, t, t, , , (c), ,  I2 , , , ,  H2 , , t, t, (d) none of these., , , ,   HI, 2t, (1997), , 4.2 Factors Influencing Rate of a Reaction, 11., , Mechanism of a hypothetical reaction,, X2 + Y2, 2XY, is given below :, (i) X2, X + X (fast), (ii) X + Y2  XY + Y (slow), (iii) X + Y, XY (fast), The overall order of the reaction will be, (a) 2, (b) 0, (c) 1.5, (d) 1, (NEET 2017), , 12. The decomposition of phosphine (PH3) on tungsten, at low pressure is a first-order reaction. It is because, the, (a) rate is proportional to the surface coverage, (b) rate is inversely proportional to the surface, coverage, (c) rate is independent of the surface coverage, (d) rate of decomposition is very slow., (NEET-II 2016), , www.neetujee.com, , 13. The rate constant of the reaction A, B is, 0.6 × 10–3 mol L–1 s–1. If the concentration of A is, 5 M, then concentration of B after 20 minutes is, (a) 3.60 M, (b) 0.36 M, (c) 0.72 M, (d) 1.08 M, (2015), 14. For a reaction between A and B the order with, respect to A is 2 and the order with respect to B is 3., The concentrations of both A and B are doubled, the, rate will increase by a factor of, (a) 12, (b) 16, (c) 32, (d) 10, (Karnataka NEET 2013), 15. In a reaction, A + B, product, rate is doubled, when the concentration of B is doubled, and rate, increases by a factor of 8 when the concentration of, both the reactants (A and B) are doubled, rate law, for the reaction can be written as, (a) rate = k[A][B]2, (b) rate = k[A] 2[B]2, (c) rate = k[A][B], (d) rate = k[A]2[B], (2012), 16. Which one of the following statements for the order, of a reaction is incorrect?, (a) Order can be determined only experimentally., (b) Order is not influenced by stoichiometric, coefficient of the reactants., (c) Order of a reaction is sum of power to the, concentration terms of reactants to express the, rate of reaction., (d) Order of reaction is always whole number., (2011), 17. The unit of rate constant for a zero order reaction is, (a) mol L–1 s–1, (b) L mol–1 s–1, 2, –2 –1, (c) L mol s, (d) s–1, (Mains 2011), 18. During the kinetic study of the reaction,, 2A + B  C + D, following results were obtained :, Run, , [A]/, mol L–1, , [B]/, mol L–1, , I., II., III., IV., , 0.1, 0.3, 0.3, 0.4, , 0.1, 0.2, 0.4, 0.1, , Initial rate of, formation of, D/mol L–1 min–1, 6.0 × 10–3, 7.2 × 10–2, 2.88 × 10–1, 2.40 × 10–2, , Based on the above data which one of the following, is correct?, (a) Rate = k[A]2[B], (b) Rate = k[A][B], (c) Rate = k[A]2[B]2, (d) Rate = k[A][B]2, (2010), , www.mediit.in

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Chemical Kinetics, , 29, , 19. For the reaction, A + B products, it is observed, that, (i) on doubling the initial concentration of A only,, the rate of reaction is also doubled and, (ii) on doubling the initial concentration of both A, and B, there is a change by a factor of 8 in the, rate of the reaction., The rate of this reaction is given by, (a) rate = k[A][B]2, (b) rate = k[A]2[B]2, (c) rate = k[A][B], (d) rate = k[A]2[B] (2009), 20. The bromination of acetone that occurs in acid, solution is represented by this equation., CH3COCH3(aq) + Br2(aq), CH3COCH2Br(aq), +, –, + H (aq) + Br (aq), These kinetic data were obtained for given reaction, concentrations., Initial concentrations, M, [CH3COCH3], [Br2], [H+], 0.30, 0.05, 0.05, 0.30, 0.10, 0.05, 0.30, 0.10, 0.10, 0.40, 0.05, 0.20, Initial rate, disappearance of Br2, M s–1, 5.7 × 10–5, 5.7 × 10–5, 1.2 × 10–4, 3.1 × 10–4, Based on these data, the rate equation is, (a) Rate = k [CH3 COCH 3][Br 2][H +]2, (b) Rate = k [CH3COCH 3][Br 2][H +], (c) Rate = k [CH3 COCH 3][H+], (d) Rate = k [CH3COCH3][Br2], (2008), 21. The reaction of hydrogen and iodine monochloride, is given as :, H2(g) + 2ICl(g), 2HCl(g) + I2(g), This reaction is of first order with respect to H2(g), and ICl(g), following mechanisms were proposed., Mechanism A :, H2(g) + 2ICl(g), 2HCl(g) + I2(g), Mechanism B :, H2(g) + ICl(g), , HCl(g) + HI(g) ; slow, , HI(g) + ICl(g), HCl(g) + I2(g) ; fast, Which of the above mechanism(s) can be consistent, with the given information about the reaction?, (a) A and B both, (b) Neither A nor B, (c) A only, (d) B only, (2007), , www.neetujee.com, , 22. The rate of reaction between two reactants A and, B decreases by a factor of 4 if the concentration of, reactant B is doubled. The order of this reaction, with respect to reactant B is, (a) 2, (b) –2, (c) 1, (d) –1, (2005), 23. If the rate of the reaction is equal to the rate constant,, the order of the reaction is, (a) 0, (b) 1, (c) 2, (d) 3, (2003), 24. 2A B + C, It would be a zero order reaction when, (a) the rate of reaction is proportional to square of, concentration of A, (b) the rate of reaction remains same at any, concentration of A, (c) the rate remains unchanged at any concentration, of B and C, (d) the rate of reaction doubles if concentration of B, is increased to double., (2002), 25. For the reaction; 2N2O5, 4NO2 + O2, rate and rate constant are 1.02 × 10–4 and, 3.4 × 10–5 sec–1 respectively, then concentration of, N2O5 at that time will be, (a) 1.732, (b) 3, (c) 1.02 × 10–4, (d) 3.4 × 105, (2001), 26. The experimental data for the reaction,, 2A + B2, 2AB is, Experiment [A], [B2], Rate (mole s–1), 1, 0.50, 0.50 1.6 × 10–4, 2, 0.50, 1.00 3.2 × 10–4, 3, 1.00, 1.00 3.2 × 10–4, The rate equation for the above data is, (a) rate = k [A] 2[B] 2, (b) rate = k [A]2[B], (c) rate = k [B2], (d) rate = k [B2]2 (1997), 27. The given reaction,, 2FeCl3 + SnCl2, 2FeCl2 + SnCl4, is an example of, (a) third order reaction, (b) first order reaction, (c) second order reaction, (d) none of these., , (1996), , 28. The data for the reaction A + B C, is, Exp. [A]0 [B]0, Initial rate, 1 0.012 0.035, 0.10, 2, 0.024 0.070, 0.80, 3, 0.024 0.035, 0.10, 4, 0.012 0.070, 0.80, The rate law corresponds to the above data is, (a) rate = k[A][B]3, (b) rate = k[A] 2[B]2, 3, (c) rate = k[B], (d) rate = k[B]4., (1994), , www.mediit.in

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30, , 4.3 Integrated Rate Equations, 29. The rate constant for a first order reaction is, 4.606 × 10–3 s–1. The time required to reduce 2.0 g of, the reactant to 0.2 g is, (a) 100 s, (b) 200 s, (c) 500 s, (d) 1000 s (NEET 2020), 30. If the rate constant for a first order reaction is k, the, time (t) required for the completion of 99% of the, reaction is given by, (a) t = 2.303/k, (b) t = 0.693/k, (c) t = 6.909/k, (d) t = 4.606/k, (NEET 2019), 31. A first order reaction has a rate constant of, 2.303 × 10–3 s–1. The time required for 40 g of this, reactant to reduce to 10 g will be, [Given that log10 2 = 0.3010], (a) 230.3 s, (b) 301 s, (c) 2000 s, (d) 602 s, (Odisha NEET 2019), 32. The correct difference between first and second, order reactions is that, (a) the rate of a first-order reaction does not, depend on reactant concentrations; the rate of a, second-order reaction does depend on reactant, concentrations, (b) the half-life of a first-order reaction does not, depend on [A]0 ; the half-life of a second-order, reaction does depend on [A]0, (c) a first-order reaction can be catalysed; a secondorder reaction cannot be catalysed, (d) the rate of a first-order reaction does depend on, reactant concentrations; the rate of a secondorder reaction does not depend on reactant, concentrations., (NEET 2018), 33. When initial concentration of the reactant is, doubled, the half-life period of a zero order reaction, (a) is halved, (b) is doubled, (c) is tripled, (d) remains unchanged., (NEET 2018), , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 36. When initial concentration of a reactant is doubled, in a reaction, its half-life period is not affected. The, order of the reaction is, (a) second, (b) more than zero but less than first, (c) zero, (d) first. (2015, Cancelled), 37. A reaction is 50% complete in 2 hours and 75%, complete in 4 hours. The order of reaction is, (a) 1, (b) 2, (c) 3, (d) 0, (Karnataka NEET 2013), 38. The half-life of a substance in a certain enzymecatalysed reaction is 138 s. The time required for, the concentration of the substance to fall from, 1.28 mg L–1 to 0.04 mg L–1 is, (a) 414 s, (b) 552 s, (c) 690 s, (d) 276 s (Mains 2011), 39. Half-life period of a first order reaction is 1386, seconds. The specific rate constant of the reaction is, (a) 0.5 × 10–2 s–1, (b) 0.5 × 10–3 s–1, (c) 5.0 × 10–2 s–1, (d) 5.0 × 10–3 s–1 (2009), 40. If 60% of a first order reaction was completed, in 60 minutes, 50% of the same reaction would, be completed in approximately (log 4 = 0.60,, log 5 = 0.69), (a) 45 minutes, (b) 60 minutes, (c) 40 minutes, (d) 50 minutes. (2007), 41. In a first-order reaction, A B, if k is rate constant, and initial concentration of the reactant A is 0.5 M,, then the half-life is, log2, log 2, (a), (b), k, k 0.5, 0.693, ln2, (c) k, (d) 0.5, (2007), k, 42. For a first order reaction A, B the reaction rate, at reactant concentration of 0.01 M is found to be, 2.0 × 10–5 mol L–1 s–1. The half-life period of the, reaction is, (a) 30 s, (b) 220 s, (c) 300 s, (d) 347 s, (2005), , 34. A first order reaction has a specific reaction rate of, 10–2 sec–1. How much time will it take for 20 g of the, reactant to reduce to 5 g?, (a) 138.6 sec, (b) 346.5 sec, (c) 693.0 sec, (d) 238.6 sec (NEET 2017), , 43. The rate of a first order reaction is, 1.5 × 10–2 mol L–1 min–1 at 0.5 M concentration of, the reactant. The half-life of the reaction is, (a) 0.383 min, (b) 23.1 min, (c) 8.73 min, (d) 7.53 min, (2004), , 35. The rate of first-order reaction is 0.04 mol L–1 s–1 at, 10 seconds and 0.03 mol L–1 s–1 at 20 seconds after, initiation of the reaction. The half-life period of the, reaction is, (a) 44.1 s, (b) 54.1 s, (c) 24.1 s, (d) 34.1 s (NEET-I 2016), , 44. The reaction A B follows first order kinetics. The, time taken for 0.8 mole of A to produce 0.6 mole of, B is 1 hour. What is the time taken for conversion of, 0.9 mole of A to produce 0.675 mole of B ?, (a) 1 hour, (b) 0.5 hour, (c) 0.25 hour, (d) 2 hours, (2003), , www.neetujee.com, , www.mediit.in

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Chemical Kinetics, , , , 31, , 45. For a first-order reaction, the half-life period is, independent of, (a) first power of final concentration, (b) cube root of initial concentration, (c) initial concentration, (d) square root of final concentration., (1999), , 4.4 Temperature Dependence of the Rate of, a Reaction, 46. For a reaction, activation energy Ea = 0 and the rate, constant at 200 K is 1.6 × 106 s–1. The rate constant at, 400 K will be, [Given that gas constant R = 8.314 J K–1 mol–1], (a) 3.2 × 104 s–1, (b) 1.6 × 106 s–1, 3 –1, (c) 1.6 × 10 s, (d) 3.2 × 106 s–1, (Odisha NEET 2019), 47. The addition of a catalyst during a chemical reaction, alters which of the following quantities?, (a) Enthalpy, (b) Activation energy, (c) Entropy, (d) Internal energy, (NEET-I 2016), 48. The activation energy of a reaction can be, determined from the slope of which of the following, graphs?, 1, T, 1, (a) lnk vs, vs, (b), ln k T, T, ln k, vs T, (c) ln k vs T, (d), T, (2015, Cancelled), 49. What is the activation energy for a reaction if its rate, doubles when the temperature is raised from 20 °C, to 35 °C?, (R = 8.314 J mol–1 K–1), –1, (a) 34.7 kJ mol, (b) 15.1 kJ mol–1, (c) 342 kJ mol–1, (d) 269 kJ mol–1, (NEET 2013), 50. In a zero-order reaction, for every 10 °C rise of, temperature, the rate is doubled. If the temperature, is increased from 10 °C to 100 °C, the rate of the, reaction will become, (a) 256 times, (b) 512 times, (c) 64 times, (d) 128 times. (2012), 51. Activation energy (Ea) and rate constants (k1 and k2), of a chemical reaction at two different temperatures, (T1 and T2) are related by, (a) ln, , k2, , , , k1, k2, , , , , , (b) ln, , www.neetujee.com, , k1, , Ea  1 1 , , , R  T1 T2 , Ea  1, 1 , , , , , , R T2  T , 1, , , , k2, , , , Ea  1, , , , 1 , , , R T2 T1 , , , (d) ln k2  Ea 1  1 , k, R T T , , (c), , , ln, , k1, , (Mains 2012), , , , 1, , 1, , 2, , 52. The rate of the reaction,, 2NO + Cl2 2NOCl is given by the rate equation,, rate = k[NO]2[Cl2]. The value of the rate constant, can be increased by, (a) increasing the temperature, (b) increasing the concentration of NO, (c) increasing the concentration of the Cl2, (d) doing all of these., (Mains 2010), 53. The rate constants k1 and k2 for two different reactions, are 1016· e–2000/T and 1015 · e–1000/T, respectively. The, temperature at which k1 = k2 is, 1000, K, (a) 2000 K, (b), 2.303, 2000, (c) 1000 K, (d), K, (2008), 2.303, 54. The temperature dependence of rate constant (k) of, a chemical reaction is written in terms of Arrhenius, equation, k = A×e–E*/RT. Activation energy (E*) of, the reaction can be calculated by plotting, 1, (a) k vs T, (b) k vs, log T, 1, 1, (c) log k vs, (d) log k vs, T, logT (2003), 55. The activation energy for a simple chemical reaction, A B is Ea in forward direction. The activation, energy for reverse reaction, (a) is negative of Ea, (b) is always less than Ea, (c) can be less than or more than Ea, (d) is always double of Ea., (2003), 56. When a biochemical reaction is carried out in, laboratory, outside the human body in absence of, enzyme, then rate of reaction obtained is 10–6 times,, the activation energy of reaction in the presence of, enzyme is, (a) 6/RT, (b) P is required, (c) different from Ea obtained in laboratory, (d) can’t say anything., (2001), 57. How enzymes increases the rate of reactions?, (a) By lowering activation energy, (b) By increasing activation energy, (c) By changing equilibrium constant, (d) By forming enzyme substrate complex, , (2000), , www.mediit.in

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32, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 58. Activation energy of a chemical reaction can be, determined by, (a) evaluating rate constants at two different, temperatures, (b) evaluating velocities of reaction at two different, temperatures, (c) evaluating rate constant at standard temperature, (d) changing concentration of reactants., (1998), , (a) does not change, (c) decreases, , 59. By the action of enzymes, the rate of biochemical, reaction, , (b) increases, (d) either (a) or (c)., (1994), , 4.5 Collision Theory of Chemical Reactions, 60. An increase in the concentration of the reactants of, a reaction leads to change in, (a) activation energy (b) heat of reaction, (c) threshold energy, (d) collision frequency., (NEET 2020), , ANSWER KEY, , 1., , (d), , 2., , (b), , 3., , (b), , 4., , (d), , 5., , (d), , 6., , (c), , 7., , (b), , 8., , (b), , 9., , (a), , 10., , (b), , 11., , (c), , 12., , (a), , 13., , (c), , 14., , (c), , 15., , (d), , 16., , (d), , 17., , (a), , 18., , (d), , 19., , (a), , 20., , (c), , 21., , (d), , 22., , (b), , 23., , (a), , 24., , (b), , 25., , (b), , 26., , (c), , 27., , (a), , 28., , (c), , 29., , (c), , 30., , (d), , 31., , (d), , 32., , (b), , 33., , (b), , 34., , (a), , 35., , (c), , 36., , (d), , 37., , (a). 38., , (c), , 39., , (b), , 40., , (a), , 41., , (c), , 42., , (d), , 43., , (b), , 44., , (a), , 45., , (c), , 46., , (b), , 47., , (b), , 48., , (a), , 49., , (a), , 50., , (b), , 51., , (b,d) 52., , (a), , 53., , (b), , 54., , (c), , 55., , (c), , 56., , (c), , 57., , (a), , 58., , (a), , 59., , (b), , 60., , (d), , Hints & Explanations, 1. (d) : For the given chemical reaction,, d[N2], 1 d[H2] 1 d[NH ], , Rate of reaction , , 3, 2 dt, dt, 3 dt, 2. (b) : For the reaction, 2N2O5  4NO2 + O2, , 1, 2, , d[O2 ], 1 d[NO2 ], 1 d[N2O5 ], , , 2 dt, 4 dt, dt, k, , 1, , k, , 1, , k, , k  k  k, k  2k; k  k, 4, 2, , 3. (b) : N2O5(g)  2NO2(g) + 1/2O2(g), For the given reaction the rate is written as, d[N2O5] 1 d[NO2 ] 2d[O2], , , dt, 2 dt, dt, d[N2O5 ],  6.25  103 mol L1 s1, Given that, dt, d[NO ], 2, ,  2  6.25  103 = 1.25 × 10–2 mol L–1 s–1, dt, 3, d[O ] 6.25  10, 2, and, ,  3.125  103 mol L1s1, dt, 2, 4. (d) : For reaction, N2 + 3H2  2NH3, d[N2 ], 1 d[NH3], 1 d[H2 ], Rate , , , 2 dt, 3 dt, dt, www.neetujee.com, , Given,, , d[NH3 ], ,  2 104 mol L1 s1, , dt, d[H2 ] 3 d[NH3 ] 3, , ,   2 104, dt, 2 dt, 2, d[H2 ], 4, ,  3 10 mol L1 s1, dt, 5. (d) : For the given reaction,, BrO3–(aq) + 5Br–(aq) + 6H+ (aq)  3Br2(l) + 3H2 O(l), Rate of reaction in terms of Br2 and Br– is,, 1 d[Br2 ], 1 d[Br ], Rate , , 3 dt, 5 dt, d[Br2 ], 3 d[Br ], , , dt, 5 dt, 6. (c) : N2(g) + 3H2(g)  2NH3(g), d[N2 ], d[H2 ], d[NH3 ], Rate , , , dt, 3dt, 2dt, d[NH3], 2 d[H2 ], , Hence, , dt, 3 dt, (b) : 2A + B  3C + D, d[A], d[B] d[C] d[D], rate , , , , 2dt, dt, 3dt, dt, Negative sign shows the decrease in concentration., 7., , www.mediit.in

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Chemical Kinetics, , 33, , 8. (b) : 3A  2B, 1 d[B], 1 d[A], Rate of the reaction =, , 2 dt, 3 dt, d[B], 2 d[A], , , dt, 3 dt, 1 d[Br ], 1 d[Br2 ], 9. (a) : Rate of reaction =, , 3 dt, 5 dt, d[Br ], 3 d[Br2 ], , , dt, 5 dt, 10. (b) : For H2(g) + I2(g)  2HI(g), the rate of reaction is, [H2 ], [I ] 1 [HI], ,  2 , t, t, 2 t, Negative sign shows disappearance of reactant and, positive sign shows the appearance of product., 11. (c) : Note : Correct the reactions given in question, as, X2, X+ X, (fast), X + Y2, XY + Y, (slow), Slow step is the rate determining step., ...(i), Rate = k[X][Y2], 2, [X], Equilibrium constant for fast step, K =, [X2 ], [X] = K[X2 ], By substituting [X] in equation (i), we get, Rate = k, , (2 y)b 2R, b, = 21, (y) b  R  (2)  2, Thus, b = 1, From data of (iii) experiment,, (2x)a(2y)b = 8R, From eqn.a (v) and (vi),, (2x) 8R, ,  (2)a  4 = 22, a, 2R, (x), , Thus, a = 2. By replacing the values of a and b in rate law;, rate = k[A]2 [B], 16. (d) : Order of a reaction is not always whole number., It can be zero, or fractional also., 17. (a) : Rate = k[A]0, mol L–1 s–1 = k, Thus, the unit of rate constant is mol L–1 s–1., 18. (d) : Let the rate of reaction be given by :, rate = k[A]a[B]b., Now consider2II and III where [A] is constant., a, b, 7.2  10, [0.3] [0.2], 2.88  101 [0.3]a[0.4]b, , ...(i), ...(ii), ...(iii), ...(iv), ...(v), , , , a, ,  1,  , a1, 4 4 , Rate = k[A][B]2, 1, , 2, , 12. (a) : At low pressure, rate is proportional to the, surface coverage and is of first order while at high, pressure, it follows zero order kinetics due to complete, coverage of surface area., 13. (c) : Reaction is of zero order as the unit of rate, constant is mol L–1 s–1., Concentration of B = k × t, = 0.6 × 10–3 × 20 × 60 = 0.72 M, 14. (c) : Rate1 = k[A]2 [B]3, Rate2 = k[2A]2[2B]3, Rate2 = 32k[A]2[B]3,  Rate2 = 32(Rate1), , www.neetujee.com, , , , 6.0  103 [0.1]a[0.1]b, 2.4  102 [0.4]a[0.1]b, ,  Order of reaction = 1   1.5, , 15. (d) : [A] [B] Rate, x, y, R, x, 2y, 2R, 2x, 2y, 8R, Let the rate law ; rate = k[A]a [B]b, From data given, (x)a(y)b = R, (x)a(2y)b = 2R, Dividing eqn. (v) by (iv),, , b, , 1,     b  2, 2 , 4, Now consider I and IV,, 1, , 1/2, K[X2 ] [Y2] = k1[X2] 3[Y2], , 2, , ...(vi), , 19. (a) : R = k[A]m[B]n, ...(i), 2R = k[2A]m[B]n, ...(ii), 8R = k[2A]m[2B]n, ...(iii), from (i), (ii) and (iii), m = 1, n = 2, So, rate = k[A][B]2, 20. (c) : From the first two experiments, it is clear that, when concentration of Br 2 is doubled, the initial rate, of disappearance of Br 2 remains unaltered. So, order of, reaction with respect to Br 2 is zero. Thus, the probable, rate law for the reaction will be : k[CH 3COCH 3][H +], 21. (d) : The slow step is the rate determining step and, it involves 1 molecule of H2(g) and 1 molecule of ICl(g)., Hence, the rate will be,, r = k[H2(g)] [ICl(g)], i.e., the reaction is 1st order with respect to H2(g) and ICl(g)., 22. (b) : Rate of reaction = k [A] [B],   order of reaction w.r.t. A,   order of reaction w.r.t. B, , www.mediit.in

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34, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , , [B], , r1 = k[A], r2 = r1/4 = k[A] [2B], , , , , 0.80  0.070 , , 0.10 0.035 , ,  1  , , r1  k [A] [B]  , , 4      2, , , 2 , , r2 k [A] [2B], , 23. (a) : A  products, dx, dx, If   k , it means   k [A]0  k, dt, dt, Hence, order of reaction must be zero., 24. (b) : 2A  B + C, The rate equation of this reaction may be expressed as, r = k [A] 0, Order = 0, r = k,  The rate is independent of concentration of the, reactant A., 25. (b) : 2N2O5  4NO2 + O2, This is a first order reaction.,  rate = k [N2O5] ;, 1.02 104, [N O ] = rate/k =, 3, 2 5, 3.4 105, 26. (c) : For the reaction, 2A + B2  2AB,, Rate  [A]x[B2]y., On substituting the given data, we get, From experiment 1,, 1.6 × 10–4  [0.50]x [0.50]y, ...(i), From experiment 2,, 3.2 × 10–4  [0.50]x [1.00]y, ...(ii), From experiment 3,, 3.2 × 10–4  [1.00]x [1.00]y, ...(iii), On dividingx equation (iii) by (ii), we get ,, 1.00 , 1,  1 = 2x  20 = 2x  x = 0,  0.50 , , , Now, divide equation (ii) by equation (i) we get,, y, 1.00 , 2,  2 = 2y  y = 1, , ,  0.50 , Thus, rate equation is :, Rate = k[A]0 [B2]1 = k[B2], 27. (a) : For a general reaction,, xA + yB + zC  product, the order of reaction is x + y + z., Since three molecules undergo change in concentration,, therefore it is a third order reaction., 28. (c) : A + B  C, Let rate = k[A]x [B]y, where order of reaction is (x + y)., Putting the values of exp. 1, 2, and 3, we get following, equations., 0.10 = k [0.012]x [0.035]y, ...(i), 0.80 = k [0.024]x [0.070]y, ...(ii), 0.10 = k [0.024]x [0.035]y, ...(iii), Dividing (ii) by (iii), we get, , www.neetujee.com, , y, ,  2y = 8, , y= 3, , Keeping [A] constant, [B] is doubled, rate becomes 8, times., Dividing eq. (iii) by eq. (i), we get, x, 0.10  0.024 , x, , , 2, =, 1x=0, , , 0.10 0.012 , Keeping [B] constant, [A] is doubled, rate remains, unaffected. Hence, rate is independent of [A]., rate  [B]3., 29. (c) : For a first order reaction,, [R]0, 2.303, k, log, t, [R], , 3,  2  2.303  10, 2.303,  t ,  500 s, 4.606  103 1 log   4.606, 0.2, s, 30. (d) : For 1st order reaction,, 2.303, a, 2.303, 100, log, t, log, , k, 100  99, k, ax, 2.303, 2.303, 4.606, log102 ,  2  log10 , k, k, k, [A]0, 2.303, log, 31. (d) : For a first order reaction, k , , , 2.303, , 40, , t, , [A]t, , 2.303  10 , log, t, 10, 1, 2, 2, t  log 22  log 2 =  0.3010 = 602 s, 103, 103, 103, 0.693, which is, 32. (b) : For the first order reaction, t1/2 , k, independent of initial concentration [A]0., 1, For second order reaction, t1/2 , k[A]0, Half-life depends on initial concentration of reactant., [A]0, 33. (b) : (t1/2 )zero , 2k, As the half-life of a zero order reaction is directly, proportional to initial concentration.,  If [A]0 = doubled then, t1/2 = doubled., 3, , 34. (a) : For a first order reaction,, 2.303 [A]0, 20, 2 2.303, k, log, or, 10 , log, t, [A]t, t, 5, 2 2.303  0.6020, 10 , or t = 138.6 sec, t, 35. (c) : For a first order reaction, A  Products and, for concentration of the reactant at two different times,, , www.mediit.in

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Chemical Kinetics, k, , 2.303, , log, , 35, , t1/2 , , t2  t1, [A]2, (rate)1, 2.303, k, log, t2  t1, (rate)2, 2.303,  0.04 , 1,  0.0287 sec, k, log , 0.03 , (20 10), 0.693, 0.693, t1/2 , ,  24.14 sec, k, 0.0287 sec1, , (Q rate  [A]), , 0.5, For first order reaction,, 0.693 0.693  0.5, t1/2 , ,  23.1 min, k, 1.5 102, , 36. (d) : Half-life period of a first order reaction is, independent of initial concentration,, 0.693, t1/2 , k, , 44. (a) : In case I, In case II, AB, A  B, 0.8, 0, 0.9, 0, 0.2, 0.6, 0.225 0.675, 1, 3, 1, 3, The time taken for the completion of same fraction of, change is independent of initial concentration., 45. (c) : For the first order reaction, rate constant is, 1 a, given by, k 1 ln, t a x, a = initial concentration, (a – x) = concentration at t time, At t = t1/2, x = a/2, 1, a, 1,  k1 , ln,  k1  ln 2, t1/2 a  a / 2, t1/2, 0.693,  k1 , t1/2, , 37. (a) : As t75% = 2 × t50%, the order of the reaction is, one., 38. (c) : Fall of concentration from 1.28 mg L–1, to 0.04 mg L–1 requires 5 half-lives.,  Time required = 5 × t1/2 = 5 × 138 = 690 s, 39. (b) : Given, t1/2 = 1386 s, For a first order reaction,, 0.693 (k  rate constant), t1/2 , k, 0.693,  1386 ,  k = 5 × 10–4 s–1 = 0.5 × 10–3 s–1, k, 40. (a) : For a first order reaction,, 2.303, a, k, log, t, a x, 2.303 100 2.303, k, log, ,  log 2.5  0.0153, 60, 40, 60, 2.303, 100 2.303, log, ,  log 2 = 45.31 min., Again, t , 1/2, 50, 0.0153, k, 41. (c) : For a 1st order reaction,, a, 2.303, k  t log10 a  x, 2.303, a, At t, ,k, log, 10, 1/2, a, t1/2, a, 2, or t1/2  2.303 log10 2  ln2, k, k, 42. (d) : A, B, Rate of reaction = 2 × 10–5 mol L–1 s–1,  order of reaction is n = l, rate = k [A]n = k[A], k is the rate constant., [A] = 0.01 M, 2 105, 0.693, k,  2 103 , k , t1/2, 0.01, , www.neetujee.com, , 0.693, , = 346.5  347 s, 2 103,  dx ,  kC, 43. (b) : Rate,  , 2, dt, i.e., 1.5 × 10–2 = k × 0.5 or, k  1.5 10, , [A]1, , , , Therefore, t1/2 is independent of initial concentration., 46. (b) : According to Arrhenius equation,, k, log 2  Ea T,  2  T1 , k1 2.303 R TT ,  2 1 , k2, k2, 1, log,  0;, 6, 6, 1.6 10, 1.6 10, , , , , , , k2 = 1.6 × 106 s–1, 47. (b) : A catalyst provides an alternate path to the, reaction which has lower activation energy., 48. (a) : According to Arrhenius equation,, k = Ae–E a/RT, Ea, Slope – Ea, ln k, ln k  ln A  RT, R, Hence, if ln k is plotted, , 1/T, , against 1/T, slope of the line will be , 49. (a) : log, , k2, k1, , , , Ea, , Ea, , ., , R, ,  1 1 , , 2.303 R T1 T2 , , k2 = 2k1, T1 = 20 + 273 = 293 K, or T2 = 35 + 273 = 308 K, , www.mediit.in

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36, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, –1, , 2000 1000, 2.303 , , T, T, 1000, or,  2.303 or T  1000 K, T, 2.303, 54. (c) : On plotting log k vs 1/T, we get a straight line,, the slope indicates the value of activation energy., , –1, , R = 8.314 J mol K, Ea,  1 1 , log 2 , , , , 2.303  8.314 293 308 , E, 0.3010  a, 15, , 19.147 293  308, Ea = 34673 J mol–1 = 34.7 kJ mol–1, 50. (b) : At 10°C rise, rate increases by 2., , or, , 55. (c) : Activation energy is the minimum amount, of energy required to convert reactant into product., Activation energy for reverse reaction can be less than, or more than Ea depending whether the reaction is, exothermic or endothermic., , 10010 , , , r100C 10, 2, r10C, ,  29  512 times, , 51. (b, d) : k1 = Ae–Ea/RT1, k2 = Ae–Ea/RT2, ln k1 = ln A – Ea/RT1, ln k2 = ln A – Ea/RT2, From eq.(i) and (ii), we have, E, E, ln k2  ln k1  ln A  a  ln A  a, RT2, RT1, , ...(i), ...(ii), , 57. (a) : Enzymes act like catalyst in biochemical, reactions. Presence of an enzyme increases the rate of, reaction by lowering the activation energy of the reactant., , k2 Ea  1, 1 , ,  , k1 R  T1 T2 , k2, Ea  1 1 ,  ln     , k1, R  T2 T1 ,  ln, , , , 56. (c) : According to k = Ae –Ea /RT (Arrhenius equation),, the activation energy of a reaction in the presence of, enzyme is different from Ea obtained in laboratory., , 52. (a) : Rate constant is independent of the initial, concentration of the reactants. It has a constant value at, fixed temperature. According to Arrhenius equation, the, value of rate constant can be increased by increasing the, temperature., 53. (b) : k1 = 1016 e–2000/T, k2 = 1015 e–1000/T, When, k1 = k 2, 1016 e–2000/T = 1015e–1000/T, , 58. (a) :kAccording, Ea to Arrhenius equation :, T2  T1 , log 2 , , , k, 2.303R T T , 1,  2 1 , where Ea = activation energy, R = gas constant = 8.314 J K–1 mol–1, k1 and k2 are rate constants of the reaction at two different, temperatures T1 and T2 respectively., 59. (b) : Since the enzymes are regarded as biological, catalysts, therefore their action increases the rate of, biological reaction., , or 10e–2000/T = e–1000/T, Taking natural logarithm of both sides, we get, 2000 1000, ln 10 , , T, T, , 60. (d) : Collision frequency  no. of reacting molecules, or atoms, Higher the concentration of reactant molecules, higher is, the probability of collision and so the collision frequency., , , , , , , , , , , , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 5, , 5.1, 1., , 2., , Surface Chemistry, , Adsorption, , The correct option representing a Freundlich, adsorption isotherm is, x, x, (a), (b),  kp0.3,  kp2.5, m, m, x, 0.5, x, (c)  kp, (d)  kp1, m, m, (Odisha NEET 2019), Which one of the following characteristics is, associated with adsorption?, (a) G and H are negative but S is positive., (b) G and S are negative but H is positive., (c) G is negative but H and S are positive., (d) G, H and S all are negative., (NEET-I 2016), , 3., , In Freundlich adsorption isotherm, the value of 1/n, is, (a) between 0 and 1 in all cases, (b) between 2 and 4 in all cases, (c) 1 in case of physical absorption, (d) 1 in case of chemisorption., (2012), , 4., , If x is amount of adsorbate and m is amount of, adsorbent, which of the following relations is not, related to adsorption process?, (a) x/m = f (p) at constant T, (b) x/m = f (T) at constant p, (c) p = f (T) at constant (x/m), x, (d)  p  T, (2011), m, , 5., , The Langmuir adsorption isotherm is deduced, using the assumption, (a) the adsorption sites are equivalent in their, ability to adsorb the particles, (b) the heat of adsorption varies with coverage, (c) the adsorbed molecules interact with each other, (d) the adsorption takes place in multilayers., (2007), , www.neetujee.com, , 6., , A plot of log(x/m) versus log P for the adsorption of, a gas on a solid gives a straight line with slope equal, to, (a) log k, (b) –log k, (c) n, (d) 1/n, (2006, 1994), , 7., , Which is not correct regarding the adsorption of a, gas on surface of a solid?, (a) On increasing temperature adsorption increases, continuously., (b) Enthalpy and entropy change is negative., (c) Adsorption is more for some specific substance., (d) It is a reversible reaction., (2001), , 5.2, , Catalysis, , 8., , Which one of the following statements is not, correct?, (a) The value of equilibrium constant is changed, in the presence of a catalyst in the reaction at, equilibrium., (b) Enzymes catalyse mainly biochemical reactions., (c) Coenzymes increase the catalytic activity of, enzyme., (d) Catalyst does not initiate any reaction., (NEET 2017), , 9., , Which one of the following statements is incorrect, about enzyme catalysis?, (a) Enzymes are mostly proteinous in nature., (b) Enzyme action is specific., (c) Enzymes are denatured by ultraviolet rays and, at high temperature., (d) Enzymes are least reactive at optimum, temperature., (2012), , 10. The enzyme which hydrolyses triglycerides to fatty, acids and glycerol is called, (a) maltase, (b) lipase, (c) zymase, (d) pepsin., (2004), , www.mediit.in

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38, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 11. According to the adsorption theory of catalysis, the, speed of the reaction increases because, (a) the concentration of reactant molecules at the, active centres of the catalyst becomes high due, to adsorption, (b) in the process of adsorption, the activation, energy of the molecules becomes large, (c) adsorption produces heat which increases the, speed of the reaction, (d) adsorption lowers the activation energy of the, reaction., (2003), , I., , 5.3, , Colloids, , 12. A colloidal system has particles of which of the, following size?, (a) 10–9 m to 10–12 m (b) 10–6 m to 10–9 m, (c) 10–4 m to 10–10 m (d) 10–5 m to 10–7 m, (1996), , 5.4, , Classification of Colloids, , 13. Measuring zeta potential is useful in determining, which property of colloidal solution?, (a) Viscosity, (b) Solubility, (c) Stability of the colloidal particles, (d) Size of the colloidal particles, (NEET 2020), 14. Which mixture of the solutions will lead to, the formation of negatively charged colloidal, [AgI]I– sol ?, (a) 50 mL of 0.1 M AgNO3 + 50 mL of 0.1 M KI, (b) 50 mL of 1 M AgNO3 + 50 mL of 1.5 M KI, (c) 50 mL of 1 M AgNO3 + 50 mL of 2 M KI, (d) 50 mL of 2 M AgNO3 + 50 mL of 1.5 M KI, (NEET 2019), 15. On which of the following properties does the, coagulating power of an ion depend?, (a) The magnitude of the charge on the ion alone, (b) Size of the ion alone, (c) Both magnitude and sign of the charge on the, ion, (d) The sign of charge on the ion alone, (NEET 2018), 16. The coagulation values in millimoles per litre of the, electrolytes used for the coagulation of As 2S3 are, given below :, , www.neetujee.com, , (NaCl) = 52,, , II. (BaCl2) = 0.69,, , III. (MgSO4) = 0.22, The correct order of their coagulating power is, (a) I > II > III, (c) III > II > I, , (b) II > I > III, (d) III > I > II, (NEET-II 2016), , 17. Fog is a colloidal solution of, (a) solid in gas, (b) gas in gas, (d) gas in liquid., (c) liquid in gas, (NEET-I 2016), 18. Which property of colloidal solution is independent, of charge on the colloidal particles?, (a) Electroosmosis, (b) Tyndall effect, (c) Coagulation, (d) Electrophoresis, (2015, Cancelled 2014), 19. The protecting power of lyophilic colloidal sol is, expressed in terms of, (a) coagulation value, (b) gold number, (c) critical micelle concentration, (d) oxidation number., (2012), 20. Which one of the following forms micelles in, aqueous solution above certain concentration?, (a) Dodecyl trimethyl ammonium chloride, (b) Glucose, (c) Urea, (d) Pyridinium chloride, (2005), 21. Position of non-polar and polar part in micelle, (a) polar at outer surface but non-polar at inner, surface, (b) polar at inner surface non-polar at outer surface, (c) distributed over all the surface, (d) are present in the surface only., (2002), 22. Which one of the following method is commonly, used method for destruction of colloid?, (a) Dialysis, (b) Condensation, (c) Filteration by animal membrane, (d) By adding electrolyte, (2000), 23. At the critical micelle concentration (CMC) the, surfactant molecules, (a) associate, (b) dissociate, (c) decompose, (d) become completely soluble., (1998), , www.mediit.in

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Surface Chemistry, , 39, , 24. The ability of anion, to bring about coagulation of a, given colloid, depends upon, (a) magnitude of the charge, (b) both magnitude and charge, (c) its charge only, (d) sign of the charge alone., (1997), , 25. When a few typical solutes are separated by a, particular selective membrane such as protein, particles, blood corpuscles, this process is called, (a) transpiration, (b) endosmosis, (c) dialysis, (d) diffusion., (1996), , ANSWER KEY, , 1., 11., 21., , (a), (d), (a), , 2., 12., 22., , 3., 13., 23., , (d), (b), (d), , (a), (c), (a), , 4., 14., 24., , (d), (b), (b), , 5., 15., 25., , (a), (c), (c), , 6., 16., , (d), (c), , 7., 17., , (a), (c), , 8., 18., , (a), (b), , 9., 19., , (d), (b), , 10., 20., , (b), (a), , Hints & Explanations, 1., x, , (a) : Freundlich adsorption isotherm equation is, 1,  1, , n,  kp 1   0 ,  n, , m, , 2. (d) : As the molecules of the adsorbate are held on, the surface of the solid adsorbent, entropy decreases i.e.,, S = –ve., As G = H – TS, For the adsorption to occur, G = –ve and it is possible, only if H = –ve., 3. (a) : Freundlich adsorption isotherm :, x, 1,  k  p1/n ; 0   1, m, n, x, 4. (d) :  p  T is the incorrect relation., m, The correct relation is, x, P, amount of absorption , m T, 5. (a) : Langmuir adsorption isotherm is based on the, assumption that every adsorption site is equivalent and, that the ability of a particle to bind there is independent, of whether nearby sites are occupied or not occupied., 6. (d) :, , k, , This is according to Freundlich adsorption isotherm., , www.neetujee.com, , 7. (a) : Adsorption is the ability of a substance to, concentrate or hold gases, liquids or dissolved substances, upon its surface. Solids adsorb greater amounts of, substances at lower temperature. In general, adsorption, decreases with increase in temperature., 8. (a) : Catalyst does not change the value of, equilibrium constant as they affect forward as well as, backward reactions equally., 9. (d) : The enzyme activity rises rapidly with, temperature and becomes maximum at definite, temperature, called optimum temperature., , 10. (b) :, , 11. (d) : Adsorption causes decrease in surface energy, which appears as heat. Thus, adsorption is an exothermic, process and hence lowers the activation energy of the, reaction., 12. (b) : Particle size of colloids lies in the range of 10–6, m to 10–9 m. Particles themselves are invisible even under, the most powerful microscope., 13. (c) : Measuring zeta potential is useful in, determining stability of the colloidal particles., 14. (b) : If colloidal sol of AgI is prepared by adding KI, solution to AgNO3 till KI is in slight excess, iodide ion, (I–) will be adsorbed on the surface of AgI thereby, giving, a negative charge to the sol., AgI + I–, AgI : I–, (From KI), , Negative sol, , www.mediit.in

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40, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 15. (c) : According to Hardy-Schulze rule, the, coagulating power of an electrolyte depends on both, magnitude and sign of the charge of the effective ion or, electrolyte., 1, 16. (c) : Coagulating power , Coagulation value, Lower the coagulation value, higher is the coagulating, power so, the correct order is :, MgSO4 > BaCl2 > NaCl, (III), (II), (I), 17. (c) : Fog is an example of aerosol in which dispersed, phase is liquid and dispersion medium is gas., 18. (b) : Tyndall effect is scattering of light by colloidal, particles which is independent of charge on them., 19. (b), 20. (a) :, , 21. (a) : Micelles are the clusters or aggregates formed, in solution by association of colloids. Usually such, , molecules have a lyophobic group and a lyophilic group., The long hydrocarbon is the lyophobic portion which, tries to recede away from the solvent water and the, ionisable lyophilic group which tends to go into water, resulting into ions. As the concentration is increased the, lyophobic parts receding away from the solvent approach, each other and form a cluster. Thus, the lyophobic ends, are in the interior and lyophilic groups projecting, outward in contact with the solvent., 22. (d) : By adding electrolytes the colloidal particles, are precipitated. The electrolytes neutralise the charge, of colloids leading to their coagulation and thus, destroy, the colloid., 23. (a) : The soap concentration at which micelles, (spherical colloid molecules) first appear is called as, critical micelle concentration (CMC). At this condition,, the surfactant molecules associate with each other., 24. (b) : Both magnitude of charge and nature of, charge effect coagulation of a given colloid. Greater the, magnitude of the charge, quicker will be the coagulation., 25. (c) : Dialysis is the process of separating the, particles of colloids from the particles of crystalloids by, means of diffusion through a selective membrane placed, in water., , , , www.neetujee.com, , www.mediit.in

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, , , , , , General Principles, and Processes of, Isolation of Elements, , CHAPTER, , 6, , 6.1 Occurrence of Metals, 1., , 2., , 3., , 6.3, , Which one is malachite from the following?, (a) CuCO3.Cu(OH)2 (b) CuFeS2, (c) Cu(OH)2, (d) Fe3O4 (NEET 2019), Identify the incorrect statement., (a) The scientific and technological process used for, isolation of the metal from its ore is known as, metallurgy., (b) Minerals are naturally occurring chemical, substances in the earth’s crust., (c) Ores are minerals that may contain a metal., (d) Gangue is an ore contaminated with undesired, materials., (Odisha NEET 2019), “Metals are usually not found as nitrates in their, ores.” Out of the following two (I and II) reasons, which is/are true for the above observation?, I. Metal nitrates are highly unstable., II. Metal nitrates are highly soluble in water., (a) I is false but II is true., (b) I is true but II is false., (c) I and II are true., (d) I and II are false, (2015, Cancelled), , 4., , Which one of the following is a mineral of iron?, (a) Malachite, (b) Cassiterite, (c) Pyrolusite, (d) Magnetite, (2012), , 5., , Cassiterite is an ore of, (a) Sb, (c) Mn, , (b) Ni, (d) Sn, , (1999), , 6.2 Concentration of Ores, 6., , Sulphide ores of metals are usually concentrated by, froth floatation process. Which one of the following, sulphide ores offer an exception and is concentrated, by chemical leaching?, (a) Galena, (b) Copper pyrite, (c) Sphalerite, (d) Argentite, (2007), , www.neetujee.com, , 7., , Extraction of Crude Metal from, Concentrated Ore, , Roasting of sulphides gives the gas X as a byproduct., This is a colourless gas with choking smell of burnt, sulphur and causes great damage to respiratory, organs as a result of acid rain. Its aqueous solution is, acidic, acts as a reducing agent and its acid has never, been isolated. The gas X is, (a) CO2, (b) SO3, (c) H2S, (d) SO2, (NEET 2013), , 6.4, , Thermodynamic Principles of Metallurgy, , 8., , Considering Ellingham diagram, which of the, following metals can be used to reduce alumina?, (a) Fe, (b) Zn, (c) Mg, (d) Cu, (NEET 2018), , 9., , In the extraction of copper from its sulphide ore, the, metal is finally obtained by the reduction of cuprous, oxide with, (a) carbon monoxide (b) copper (I) sulphide, (c) sulphur dioxide (d) iron (II) sulphide., (2015, 2012), , 10. The metal oxide which cannot be reduced to metal, by carbon is, (a) Al2O3 (b) PbO (c) ZnO (d) Fe2O3, (Karnataka NEET 2013), 11. Which of the following elements is present as the, impurity to the maximum extent in the pig iron?, (a) Manganese, (b) Carbon, (c) Silicon, (d) Phosphorus, (2011, 1998), 12. The following reactions take place in the blast, furnace in the preparation of impure iron. Identify, the reaction pertaining to the formation of the slag., (a) Fe 2O3(s ) + 3CO ( g)  2Fe(l) + 3CO2(g ), (b) CaCO3(s)  CaO(s) + CO2(g), (c) CaO(s) + SiO2(s)  CaSiO3(s), (d) 2C(s) + O2(g)  2CO(g), (Mains 2011), , www.mediit.in

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42, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 13. Which of the following statements, about the, advantage of roasting of sulphide ore before, reduction is not true?, (a) The Gf° of the sulphide is greater than those for, CS2 and H2S., (b) The Gf° is negative for roasting of sulphide ore, to oxide., (c) Roasting of the sulphide to the oxide is, thermodynamically feasible., (d) Carbon and hydrogen are suitable reducing, agents for metal sulphides., (2007), , (c) displacement with Zn, (d) liquation., , 14. Nitriding is the process of surface hardening of steel, by treating it in an atmosphere of, (a) NH3, (b) O3, (c) N2, (d) H2S, (1989), , 6.5, , Electrochemical Principles of Metallurgy, , 15. Aluminium is extracted from alumina (Al 2O3) by, electrolysis of a molten mixture of, (a) Al2O3 + HF + NaAlF4, (b) Al2O3 + CaF2 + NaAlF4, (c) Al2O3 + Na3AlF6 + CaF2, (2012), , (d) Al2O3 + KF + Na3AlF6, , 16. Purification of aluminium, by electrolytic refining,, is known as, (a) Hoope’s process, (b) Baeyer’s process, (c) Hall’s process, (d) Serpeck’s process., (1999), 17. Calcium is obtained by, (a) reduction of calcium chloride with carbon, (b) electrolysis of molten anhydrous calcium, chloride, (c) roasting of limestone, (d) electrolysis of solution of calcium chloride in, H2O., (1997), , 6.6, , Oxidation Reduction, , 18. Extraction of gold and silver involves leaching with, CN– ion. Silver is later recoveredby, (a) distillation, (b) zone refining, , 6.7, , (NEET 2017), , Refining, , 19. Identify the correct statement from the following :, (a) Wrought iron is impure iron with 4% carbon., (b) Blister copper has blistered appearance due to, evolution of CO2., (c) Vapour phase refining is carried out for Nickel, by van Arkel method., (d) Pig iron can be moulded into a variety of shapes., (NEET 2020), 20. Match items of Column I with the items of Column, II and assign the correct code :, Column I, Column II, (A) Cyanide process (i) Ultrapure Ge, (B) Froth floatation (ii) Dressing of ZnS, process, (C) Electrolytic, (iii) Extraction of Al, reduction, (D), Zone refining (iv) Extraction of Au, (v) Purification of Ni, Code :, A B C, D, (a) (i) (ii) (iii) (iv), (b) (iii) (iv) (v) (i), (c) (iv) (ii) (iii) (i), (d) (ii) (iii) (i), (v), (NEET-I 2016), 21. Which of the following pairs of metals is purified by, van Arkel method?, (a) Ga and In, (b) Zr and Ti, (c) Ag and Au, (d) Ni and Fe, (2011), 22. The method of zone refining of metals is based on, the principle of, (a) greater mobility of the pure metal than that of, the impurity, (b) higher melting point of the impurity than that, of the pure metal, (c) greater noble character of the solid metal than, that of the impurity, (d) greater solubility of the impurity in the molten, state than in the solid., (2003), , ANSWER KEY, , 1., 11., 21., , (a), (b), (b), , www.neetujee.com, , 2., 12., 22., , (d), (c), (d), , 3., 13., , (a), (d), , 4., 14., , (d), (a), , 5., 15., , (d), (c), , 6., 16., , (d), (a), , 7., 17., , (d), (b), , 8., 18., , (c), (c), , 9., 19., , (b), (d), , 10., 20., , (a), (c), , www.mediit.in

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General Principles and Processes of Isolation of Elements, , 43, , Hints & Explanations, 1. (a) : Malachite : CuCO3·Cu(OH)2, 2. (d) : An ore rarely contains only a desired substance., It is usually contaminated with earthly or undesired, materials known as gangue., 3. (a) : All nitrates are soluble in water and are quite, stable as they do not decompose easily on heating., 4. (d) : Magnetite is Fe3O4 and contains upto 70% of, iron metal., 5. (d) : Cassiterite is also known as tin stone (SnO2),, an ore of tin (Sn)., 6. (d) : Leaching process involves the treatment of the, ore with a suitable reagent as to make it soluble while, impurities remain insoluble. The ore is recovered from, the solution by suitable chemical method. Argentite or, silver glance, Ag2S is an ore of silver. Silver is extracted, from argentite by the Mac-Arthur and Forest cyanide, process (leaching process)., Ag2S + 4NaCN, 2Na[Ag(CN)2] + Na2S, 2Na[Ag(CN)2] + Zn, Na2[Zn(CN)4] + 2Ag, 7. (d), 8. (c) : Any metal oxide with lower value of G°, is more stable than a metal oxide with higher G°., This implies that the metal oxide placed higher in the, Ellingham diagram can be reduced by the metal involved, in the formation of the oxide placed lower in the diagram., The relative tendency of the various metals to act as, reducing agents is :, Ca > Mg > Al > Zn > Fe > Cu., Thus, Mg being more reducing in nature, can reduce, aluminium oxide (alumina) to aluminium., 9. (b) : It is an example of auto reduction., , 14. (a) : When steel is heated in presence of NH3, iron, nitride on the surface of steel is formed which imparts a, hard coating. This process is called nitriding., 15. (c) : Electrolytic mixture contains alumina (Al2O3),, cryolite (Na3AlF6) and fluorspar (CaF2) in the ratio of, 20 : 40 : 20. Due to presence of these, conductivity of, alumina increases and fusion temperature decreases, from 2000°C to 900°C., 16. (a) : Aluminium metal obtained from Hoope’s, electrolytic refining process is about 99.9% pure. The, cell used for this process consists of three layers. The, upper layer is pure ‘Al’, acts as cathode, the middle, layer is mixture of fluorides of Al and Ba, which acts as, electrolyte. The lowest layer is impure ‘Al’, which acts as, anode. On electrolysis pure ‘Al’ is transferred from the, bottom to the top layer, through the middle layer., 17. (b) : Calcium is obtained by the electrolysis of a, fused mixture of anhydrous CaCl2 and CaF2 in a graphite, linked tank which serves as anode. The cathode is a, hollow movable iron rod which is kept cool. During, electrolysis, calcium is deposited at cathode while Cl 2 is, liberated at anode., 18. (c) : Extraction of gold and silver involves leaching, the metal with CN– and the metals silver and gold are, later recovered by displacement method., 4M (s) + 8CN – aq) + 2H2 O(aq) + O2(g), (, 4[M(CN) ]– + 4OH –, 2 (aq), (aq), 2[M(CN) ]– + Zn, 2M + [Zn(CN) ]2–, , 2Cu2O + Cu2S, 6Cu + SO2, 10. (a) : Oxides of less reactive metals (like PbO, ZnO,, Fe2O3) can be reduced by carbon. While oxides of very, reactive metals (like Al2O3) can be reduced only by the, electrolytic method., 11. (b) : Pig iron contains about 4% carbon and many, impurities such as S, Mn, P, Si, etc. in smaller amount., 12. (c) : Slag is formed by the reaction, CaO + SiO2 CaSiO3, 13. (d) : The standard free energies of formation (Gf°), of most of the sulphides are greater than those of CS2 and, H2S. Hence, neither carbon nor hydrogen can reduce, metal sulphides to metal. The standard free energies of, , 19. (d) : (a) Pig iron is impure iron with 4% carbon., (b) Blister copper has blistered appearance due to, evolution of SO2., (c) Vapour phase refining is carried out for nickel by, Mond’s process., (d) Pig iron can be moulded into a variety of shapes., 20. (c), 21. (b) : van Arkel method is used for purification of Zr, and Ti., 22. (d) : Elements which are used as semiconductors, such as Si, Ge, Ga, etc. are refined by this method, which, is based on the difference in solubility of impurities in, molten and solid state of the metal., , formation of oxides are much lower than those of SO2., Therefore, oxidation of metal sulphides to metal oxides, is thermodynamically favourable. Hence, sulphide ore is, roasted to the oxide before reduction., , 2 (aq), , (s), , (s), , 4, , (aq), , , , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 7, , 7.1, 1., , 2., , 3., , 4., , 5., , 6., 7., , 8., , Group 15 Elements, , In which of the following compounds, nitrogen, exhibits highest oxidation state?, (a) N2H4, (b) NH3, (c) N3H, (d) NH2OH, (2012), Nitrogen forms N2, but phosphorus does not form P2,, however, it converts P4, reason is, (a) triple bond present between phosphorus atom, (b) p – p bonding is weak, (c) p – p bonding is strong, (d) multiple bonds form easily., (2001), Which of the following oxides is most acidic?, (a) As2O5, (b) P2O5, (c) N2O5, (d) Sb2O5, (1999), Which of the following has the highest dipole, moment?, (a) SbH3, (b) AsH3, (c) NH3, (d) PH3, (1997), The basic character of hydrides of the, V group elements decreases in the order, (a) NH3 > PH3 > AsH3 > SbH3, (b) SbH3 > AsH3 > PH3 > NH3, (c) SbH3 > PH3 > AsH3 > NH3, (d) NH3 > SbH3 > PH3 > AsH3, (1996), Among the following oxides, the lowest acidic is, (a) As4O6, (b) As4O10, (c) P4O6, (d) P4O10, (1996), Which of the following fluorides does not exist?, (a) NF5, (b) PF5, (c) AsF5, (d) SbF5, Which one has the lowest boiling point?, (a) NH3, (b) PH3, (c) AsH3, (d) SbH3, , 7.2, 9., , The p-Block Elements, , 11. Pure nitrogen is prepared in the laboratory by, heating a mixture of, (a) NH4OH + NaCl, (b) NH4NO3 + NaCl, (c) NH4Cl + NaOH, (d) NH4Cl + NaNO2.(1991), 12. Which of the following statement is not correct for, nitrogen?, (a) Its electronegativity is very high., (b) d-orbitals are available for bonding., (c) It is a typical non-metal., (d) Its molecular size is small., (1990), , 7.3, , 4, , 7.4, (1989), , Number of electrons shared in the formation of, nitrogen molecule is, (a) 6, (b) 10, (c) 2, (d) 8, (1992), , Ammonia, , 13. Urea reacts with water to form A which will, decompose to form B. B when passed through, Cu2+, (aq) , deep blue colour solution C is formed. What, is the formula of C from the following?, (a) CuSO4, (b) [Cu(NH3)4]2+, (c) Cu(OH)2, (d) CuCO3·Cu(OH)2, (NEET 2020), 14. Aqueous solution of ammonia consists of, +, (a) H, (b) NH, OH+– and OH–. (1991), (c), NH+, (d), , (1993), , Dinitrogen, , www.neetujee.com, , 10. Nitrogen is relatively inactive element because, (a) its atom has a stable electronic configuration, (b) it has low atomic radius, (c) its electronegativity is fairly high, (d) dissociation energy of its molecule is fairly high., (1992), , 4, , Oxides of Nitrogen, , 15. Which of the following oxides of nitrogen is, paramagnetic?, (a) NO2, (b) N2O3, (c) N2O, (d) N2O5, (1994), 16. Which of the following is a nitric acid anhydride?, (a) NO, (b) NO2, (c) N2O5, (d) N2O3, (1988), , www.mediit.in

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The p-Block Elements, , 7.5, , 45, , Nitric Acid, , 17. When copper is heated with conc. HNO3 it produces, (a) Cu(NO3)2, NO and NO2, (b) Cu(NO3)2 and N2O (c) Cu(NO3)2 and NO2, (d) Cu(NO3)2 and NO, (NEET-I 2016), 18. Zn gives H2 gas with H2SO4 and HCl but not with, HNO3 because, (a) Zn act as oxidising agent when react with HNO3, (b) HNO3 is weaker acid than H2SO4 and HCl, (c) in electrochemical series Zn is above hydrogen, (d) NO 3– is reduced in preference to hydronium, ion., (2002), 19. Sugarcane on reaction with nitric acid gives, (a) CO2 and SO2, (b) (COOH)2, (c) 2HCOOH(two moles)(d) no reaction., (1992), , 7.6, , Phosphorus - Allotropic Forms, , 20. Which of the following phosphorus is the most, reactive?, (a) Scarlet phosphorus (b) White phosphorus, (c) Red phosphorus, (d) Violet phosphorus, (1999), 21. Each of the following is true for white and red, phosphorus except that they, (a) are both soluble in CS2, (b) can be oxidised by heating in air, (c) consist of the same kind of atoms, (d) can be converted into one another., (1989), , 7.7, , Phosphine, , 22. A compound ‘X’ upon reaction with H2O produces, a colourless gas ‘Y’ with rotten fish smell. Gas ‘Y’ is, absorbed in a solution of CuSO4 to give Cu3P2 as one, of the products. Predict the compound ‘X’., (a) Ca3P2, (b) NH4Cl, (c) As2O3, (d) Ca3(PO4)2, (Odisha NEET 2019), 23. PH4I + NaOH forms, (a) PH3, (c) P4O6, , 7.8, , (b) NH3, (d) P4O10, , (1991), , Phosphorus Halides, , 24. Identify the incorrect statement related to PCl from, 5, , the following :, (a) PCl5 molecule is non-reactive., (b) Three equatorial P – Cl bonds make an angle of, 120° with each other., (c) Two axial P – Cl bonds make an angle of 180°, with each other., (d) Axial P – Cl bonds are longer than equatorial, P – Cl bonds., (NEET 2019), , www.neetujee.com, , 25. PCl3 reacts with water to form, (a) PH3, (b) H3PO3, HCl, (c) POCl3, (d) H3PO4, , 7.9, , (1991), , Oxoacids of Phosphorus, , 26. Which of the following oxoacids of phosphorus has, strongest reducing property?, (a) H4P2O7, (b) H3PO3, (c) H3PO2, (d) H3PO4, (Odisha NEET 2019), 27. Which is the correct statement for the given acids?, (a) Phosphinic acid is a monoprotic acid while, phosphonic acid is a diprotic acid., (b) Phosphinic acid is a diprotic acid while, phosphonic acid is a monoprotic acid., (c) Both are diprotic acids., (d) Both are triprotic acids., (NEET-I 2016), 28. Strong reducing behaviour of H3PO2 is due to, (a) high electron gain enthalpy of phosphorus, (b) high oxidation state of phosphorus, (c) presence of two —OH groups and one P—H bond, (d) presence of one —OH group and two P—H, bonds., (2015), 29. Which of the following statements is not valid for, oxoacids of phosphorus?, (a) Orthophosphoric acid is used in the manufacture, of triple superphosphate., (b) Hypophosphorous acid is a diprotic acid., (c) All oxoacids contain tetrahedral four, coordinated phosphorus., (d) All oxoacids contain atleast one P, O unit, and one P—OH group., (2012), 30. Oxidation states of P in H4P2O5, H4P2O6, H4P2O7 are, respectively, (a) +3, +5, +4, (b) +5, +3, +4, (c) +5, +4, +3, (d) +3, +4, +5, (2010), 31. How many bridging oxygen atoms are present in, P4O10?, (a) 6, (b) 4, (c) 2, (d) 5, (Mains 2010), 32. The structural formula of hypophosphorous acid is, O, , O, , P, , P, , (a) H, , (c) H, , OH, O, , OH, , P, H, , OH, , (b) HO, , OH, , OOH, , (d) none of these., , (1997), , www.mediit.in

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46, , 33. H3PO2 is the molecular formula of an acid of, phosphorus. Its name and basicity respectively are, (a) phosphorous acid and two, (b) hypophosphorous acid and two, (c) hypophosphorous acid and one, (d) hypophosphoric acid and two., (1992), 34. Which one of the following substance is used in the, laboratory for fast drying of neutral gases?, (a) Phosphorus pentoxide (b) Active charcoal, (c) Anhydrous calcium chloride, (d) Na3PO4, (1992), 35. P2O5 is heated with water to give, (a) hypophosphorous acid, (b) phosphorous acid (c) hypophosphoric acid, (d) orthophosphoric acid., (1991), 36. Basicity of orthophosphoric acid is, (a) 2, (b) 3, (c) 4, (d) 5, (1991), 37. When orthophosphoric acid is heated to 600°C, the, product formed is, (a) PH3, (b) P2O5, (c) H3PO3, (d) HPO3, (1989), , 7.10 Group 16 Elements, 38. Which is the correct thermal stability order for H2E, (E = O, S, Se, Te and Po) ?, (a) H2Se < H2Te < H2Po < H2O < H2S, (b) H2S < H2O < H2Se < H2Te < H2Po, (c) H2O < H2S < H2Se < H2Te < H2Po, (d) H2Po < H2Te < H2Se < H2S < H2O (NEET 2019), 39. Acidity of diprotic acids in aqueous solutions, increases in the order, (a) H2S < H2Se < H2Te (b) H2Se < H2S < H2Te, (c) H2Te < H2S < H2Se (d) H2Se < H2Te < H2S, (2014), 40. Which of the following bonds has the highest energy?, (a) S–S, (b) O–O, (c) Se–Se, (d) Te–Te, (1996), , 7.11 Dioxygen, 41. Which of the following does not give oxygen on, heating?, (a) K2Cr2O7, (b) (NH4)2Cr2O7, (c) KClO3, (d) Zn(ClO3)2, (NEET 2013), 42. Which would quickly absorb oxygen?, (a) Alkaline solution of pyrogallol, (b) Conc. H2SO4, (c) Lime water, (d) Alkaline solution of CuSO4, (1991), , www.neetujee.com, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 43. Oxygen will directly react with each of the following, elements except, (a) P, (b) Cl, (c) Na, (d) S, (1989), 44. It is possible to obtain oxygen from air by fractional, distillation because, (a) oxygen is in a different group of the periodic, table from nitrogen, (b) oxygen is more reactive than nitrogen, (c) oxygen has higher b.pt. than nitrogen, (d) oxygen has a lower density than nitrogen. (1989), , 7.12 Simple Oxides, 45. Match the following :, Oxide, Nature, (A) CO, (i) Basic, (B) BaO, (ii) Neutral, (C) Al2O3, (iii) Acidic, (D) Cl2O7, (iv) Amphoteric, Which of the following is correct option?, (A) (B), (C) (D), (a) (i), (ii), (iii) (iv), (b) (ii), (i), (iv) (iii), (c) (iii) (iv), (i), (ii), (d) (iv), (iii) (ii) (i), (NEET 2020), , 7.13 Ozone, 46. The angular shape of ozone molecule (O3) consists, of, (a) 1 and 1 bond, (b) 2 and 1 bond, (c) 1 and 2 bonds, (d) 2 and 2 bonds.(2008), 47. The gases respectively absorbed by alkaline, pyrogallol and oil of cinnamon are, (a) O3, CH4, (b) O2, O3, (c) SO2, CH4, (d) N2O, O3, (1989), , 7.15 Sulphur Dioxide, 48. Nitrogen dioxide and sulphur dioxide have some, properties in common. Which property is shown by, one of these compounds, but not by the other?, (a) Is soluble in water., (b) Is used as a food preservative., (c) Forms ‘acid-rain’., (d) Is a reducing agent., (2015, Cancelled), 49. Sulphur trioxide can be obtained by which of the, following reaction?, , (a) CaSO4 + C , , (b) Fe2(SO4)3 , , , , (c) S + H2SO4 , , , (d) H2SO4 + PCl5 , , (2012), , www.mediit.in

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The p-Block Elements, , 47, , 7.16 Oxoacids of Sulphur, 50. Which of the following oxoacid of sulphur has, — O — O — linkage, (a) H2SO3, sulphurous acid, (b) H2SO4, sulphuric acid, (c) H2S2O8, peroxodisulphuric acid, (d) H2S2O7, pyrosulphuric acid, (NEET 2020), 51. Identify the correct formula of oleum from the, following :, (a) H2S2O7, (b) H2SO3, (c) H2SO4, (d) H2S2O 8, (Odisha NEET 2019), 52. In which pair of ions both the species contain S — S, bond?, (a) S O 2–, S O2–, (b) S O2– , S O 2–, 2 7, 2 8, 4 6, 2 3, (c) S O 2– , S O2–, (d) S O 2–, S O2–, 4, , 6, , 2 7, , 2, , 7, , 2, , 3, , (NEET 2017), 53. Oleum is, (a) castor oil, (c) fuming H2SO4, , (b) oil of vitriol, (d) none of these. (1991), , 7.17 Sulphuric Acid, 54. Match List I (substances) with List II (processes), employed in the manufacture of the substances and, select the correct option., List I, List II, (Substances), (Processes), (A) Sulphuric acid, (i) Haber’s process, (B) Steel, (ii) Bessemer’s, process, (C) Sodium, (iii) Leblanc process, hydroxide, (D) Ammonia, (iv) Contact process, (a) A - (i), B - (iv), C - (ii), D - (iii), (b) A - (i), B - (ii), C - (iii), D - (iv), (c) A - (iv), B - (iii), C - (ii), D - (i), (d) A - (iv), B - (ii), C - (iii), D - (i) (Mains 2010), , 7.18 Group 17 Elements, 55. Which of the following statements is not true for, halogens?, (a) All form monobasic oxyacids., (b) All are oxidizing agents., (c) All but fluorine show positive oxidation states., (d) Chlorine has the highest electron-gain, enthalpy., (NEET 2018), 56. Which one of the following orders is correct for the, bond dissociation enthalpy of halogen molecules?, (a) Br2 > I2 > F2 > Cl2, (b) F2 > Cl2 > Br2 > I2, (c) I2 > Br2 > Cl2 > F2, (d) Cl2 > Br2 > F2 > I2, (NEET-I 2016), , www.neetujee.com, , 57. The variation of the boiling points of the hydrogen, halides is in the order HF > HI > HBr > HCl., What explains the higher boiling point of hydrogen, fluoride?, (a) There is strong hydrogen bonding between HF, molecules., (b) The bond energy of HF molecules is greater, than in other hydrogen halides., (c) The effect of nuclear shielding is much reduced, in fluorine which polarises the HF molecule., (d) The electronegativity of fluorine is much higher, than for other elements in the group., (2015), 58. Among the following which is the strongest, oxidising agent?, (a) Br, (b) I, 2, (2009), (c) Cl, (d) F2, 2, , 2, , 59. Which one of the following arrangements does, not give the correct picture of the trends indicated, against it?, (a) F2 > Cl2 > Br2 > I2 : Bond dissociation energy, (b) F2 > Cl2 > Br2 > I2 : Electronegativity, (c) F2 > Cl2 > Br2 > I2 : Oxidizing power, (d) F2 > Cl2 > Br2 > I2 : Electron gain enthalpy, (2008), 60. Which one of the following orders is not in, accordance with the property stated against it?, (a) F2 > Cl2 > Br2 > I2 : Bond dissociation energy, (b) F2 > Cl2 > Br2 > I2 : Oxidising power, (c) HI > HBr > HCl > HF : Acidic property in water, (d) F2 > Cl2 > Br2 > I2 : Electronegativity, (2006), 61. Which statement is wrong?, (a) Bond energy of F2 > Cl2, (b) Electronegativity of F > Cl, (c) F is more oxidising than Cl, (d) Electron affinity of Cl > F, , (2000), , 62. Which of the following has the greatest electron, affinity?, (a) I, (b) Br, (c) F, (d) Cl, (1996), 63. Which of the following displaces Br2 from an, aqueous solution containing bromide ions?, (a) I2, (b) I3– –, (c) Cl2, (d) Cl, (1994), 64. Which of the following species has four lone pairs of, electrons?, (a) I, (b) O, (c) Cl–, (d) He, (1993), , www.mediit.in

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48, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 7.19 Chlorine, , 7.21 Oxoacids of Halogens, , 65. Match the following :, (A) Pure nitrogen, (B) Haber process, (C) Contact process, (D) Deacon’s process, , (i) Chlorine, (ii) Sulphuric acid, (iii) Ammonia, (iv) Sodium azide or, Barium azide, Which of the following is the correct option ?, (A) (B) (C) (D), (a) (iv) (iii) (ii) (i), (b) (i), (ii) (iii) (iv), (c) (ii) (iv) (i), (iii), (d) (iii) (iv) (ii) (i), (NEET 2019), 66. When Cl2 gas reacts with hot and concentrated, sodium hydroxide solution, the oxidation number, of, fromto –5, (a)chlorine, zero to changes, +1 and zero, (b) zero to –1 and zero to +5, (c) zero to –1 and zero to +3, (d) zero to +1 and zero to –3, , 74. Which of the statements given below is incorrect?, (a) O3 molecule is bent., –, (b) ONF is isoelectronic with O2N ., (c) OF2 is an oxide of fluorine., (d) Cl2O7 is an anhydride of perchloric acid., (2015), 75. The correct order of increasing bond angles in the, following species is, –, (a) Cl2O < ClO2 < ClO2–, (b) ClO2 < Cl2O < ClO, 2, , (c) Cl2O–< ClO2– < ClO2, (d) ClO < Cl O < ClO, 2, , (2012), , 67. Which of the following is used in the preparation of, chlorine?, (a) Both MnO2 and KMnO 4, (b) Only KMnO4, (c) Only MnO2, (d) Either MnO2 or KMnO4, (1999), 68. Which of the following elements is extracted, commercially by the electrolysis of an aqueous, solution of its compound?, (a) Cl, (b) Br, (c) Al, (d) Na, (1993), 69. When chlorine is passed over dry slaked lime at, room temperature, the main reaction product is, (a) Ca(ClO2)2, (b) CaCl2, (c) CaOCl2, (d) Ca(OCl)2 (1992), 70. In the manufacture of bromine from sea water, the, mother liquor containing bromides is treated with, (a) carbon dioxide (b) chlorine, (c) iodine, (d) sulphur dioxide., (1992), 71. The bleaching action of chlorine is due to, (a) reduction, (b) hydrogenation, (c) chlorination, (d) oxidation., (1991), , 7.20 Hydrogen Chloride, 72. Bleaching powder reacts with a few drops of conc., HCl to give, (a) chlorine, (b) hypochlorous acid, (c) calcium oxide, (d) oxygen., (1989), , www.neetujee.com, , 73. Among the following, the correct order of acidity is, (a) HClO2 < HClO < HClO3 < HClO4, (b) HClO4 < HClO2 < HClO < HClO3, (c) HClO3 < HClO4 < HClO2 < HClO, (d) HClO < HClO2 < HClO3 < HClO4, (NEET-I 2016, 2007, 2005), , 2, , (2010), , 2, , 76. Which one of the following oxides is expected to exhibit, paramagnetic behaviour?, (a) CO2, (b) SiO2, (c) SO2, (d) ClO2, (2005), , 7.22 Interhalogen Compounds, 77. Match the interhalogen compounds of column-I, with the geometry in column-II and assign the, correct code., Column I, Column II, (A) XX, (i) T-shape, (B) XX3, (ii) Pentagonal bipyramidal, (C) XX5, (iii) Linear, (D) XX7, (iv) Square pyramidal, (v) Tetrahedral, Code :, A B, C, D, (a) (iii) (i), (iv) (ii), (b) (v) (iv) (iii) (ii), (c) (iv) (iii) (ii) (i), (d) (iii) (iv) (i), (ii), (NEET 2017), , 7.23 Group-18 elements, 78. Match the Xenon compounds in Column-I with its, structure in Column-II and assign the correct code., Column-I, (A) XeF4, (B) XeF6, (C) XeOF4, (D) XeO3, , (i), (ii), (iii), (iv), , Column-II, pyramidal, square planar, distorted octahedral, square pyramidal, , www.mediit.in

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The p-Block Elements, (a), (b), (c), (d), , (A), (iii), (i), (ii), (ii), , (B), (iv), (ii), (iii), (iii), , 49, , (C), (i), (iii), (iv), (i), , (D), (ii), (iv), (i), (iv), (NEET 2019, NEET-I 2016), , 79. Identify the incorrect statement, regarding the, molecule XeO4., , (a), (b), (c), (d), , XeO4 molecule is square planar., There are four p - d bonds., There are four sp3 - p,  bonds., XeO4 molecule is tetrahedral., (Karnataka NEET 2013), , 80. Which compound has planar structure?, (a) XeF4, (b) XeOF2, (c) XeO2F2, (d) XeO4, , (2000), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., 61., 71., , (c), (d), (a), (a), (b), (a), (a), (d), , 2., 12., 22., 32., 42., 52., 62., 72., , (b), (b), (a), (c), (a), (a), (d), (a), , 3., 13., 23., 33., 43., 53., 63., 73., , (c), (b), (a), (c), (b), (c), (c), (d), , 4., 14., 24., 34., 44., 54., 64., 74., , (c), (d), (a), (a), (c), (d), (c), (c), , 5., 15., 25., 35., 45., 55., 65., 75., , (a), (a), (b), (d), (b), (c), (a), (d), , 6., 16., 26., 36., 46., 56., 66., 76., , (a), (c), (c), (b), (b), (d), (b), (d), , 7., 17., 27., 37., 47., 57., 67., 77., , (a), (c), (a), (d), (b), (a), (a), (a), , 8., 18., 28., 38., 48., 58., 68., 78., , (b), (d), (d), (d), (b), (d), (a), (c), , 9., 19., 29., 39., 49., 59., 69., 79., , (a), (b), (b), (a), (b), (a,d), (c), (a), , 10., 20., 30., 40., 50., 60., 70., 80., , (d), (b), (d), (a), (c), (a), (b), (a), , Hints & Explanations, 1. (c) : N2H4  2x + 4 (+1) = 0,  2x + 4 = 0,  x = –2, NH3  x + 3(+1) = 0  x = –3, N3H  3x + 1(+1) = 0,  3x + 1 = 0  x = –1/3, NH2OH  x + 2 + 1(–2) + 1 = 0,  x + 1 = 0  x = –1, Thus, highest oxidation state is –1/3., 2. (b) : For strong -bonding, p - p bonding should, be strong. In case of P, due to larger size as compared to, N-atom, p - p bonding is not so strong., 3. (c) : Among N, P, As and Sb, the former has highest, electronegativity (EN) so its oxide is most acidic., As the electronegativity value of element increases, the, acidic character of the oxide also increases., 4. (c) : Due to greater electronegativity of nitrogen,, dipole moment for NH3 is greater., 5. (a) : All the hydrides of group V elements have one, lone pair of electrons on their central atom. Therefore,, they can act as Lewis bases. The basic character of these, hydrides decreases down the group., 6. (a) : The acidic character of the oxides decreases, with the decrease in the oxidation state and also decreases, , www.neetujee.com, , down the group., 7., , (a) : Nitrogen cannot form pentahalides because, , it cannot expand its octet due to non-availability of dorbitals., 8., , (b) : Boiling pointof hydrides increases with increase, , in atomic number but ammonia has exceptionally high, boiling point due to hydrogen bonding. Thus, the correct, order of boiling point is,, BiH3 > SbH3 > NH3 > AsH3 > PH3, 9., , (a) : Nitrogen molecule is diatomic containing a, , triple bond between two N atoms, N N therefore,, nitrogen molecule is formed by sharing six electrons., 10. (d) : N2 molecule contains triple bond between N, atoms having very high dissociation energy (946 kJ mol–1), due to which it is relatively inactive., 11. (d) : NH4Cl + NaNO2, , Heat, , NH4NO2 + NaCl, , N2 + 2H2O, 12. (b) : In case of nitrogen, d-orbitals are not available, for bonding. N : 1s2 2s2 2p3, , www.mediit.in

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The p-Block Elements, , 51, , 32. (c) : The formula of hypophosphorous acid is, H3PO2 as shown in (c). It is a monobasic acid., 33. (c) : H3PO2 is named as hypophosphorous acid. As, it contains only one P—OH group, its basicity is one., , O, P, H, H, 34. (a) : P2O5 absorbs moisture much readily than, anhydrous CaCl2., HO, , Water vapour and CO2 are removed by solidification., The remaining constituents of liquid air i.e., liquid, oxygen and liquid nitrogen are separated by means, of fractional distillation as fractional distillation is a, process of separation of mixture based on the difference, in their boiling points. (b.pt. of O2 = –183°C : b.pt. of, N2 = –195.8°C)., 45. (b) : CO — neutral, BaO — basic,, Al2O3 — amphoteric and Cl2O7 — acidic., 46. (b) : The angular shape of ozone molecule (O3) :, , , 35. (d) : P2O5 + 3H2O ,  2H3PO4, 36. (b) : Orthophosphoric acid, H3PO4 contains three, P—OH groups and is therefore, tribasic., O, , O, , O, , , , OH, , OH, 37. (d) : On heating, it gives pyrophosphoric acid at, 525 K and metaphosphoric acid at 875 K., 2H PO, , 525 K, 875 K, HPO, 2HPO3, 3, 4, –H2O, 427, –H2O, Metaphosphoric, Orthophosphoric, Pyrophosphoric, acid, acid, acid, , O, , 2, , 43, , 2, , 41. (b) : (NH4)2Cr2O7  N2 + Cr2O3 + 4H2O, , Zn(ClO3)2 ,  ZnCl2 + 3O2, , , KClO3  KCl + 3/2O2, , , 2K2 Cr2O7  2K2CrO4 + Cr2 O3 + 3/2O2, 42. (a) : Alkaline solution of pyrogallol absorbs oxygen, quickly., 43. (b) : Chlorine does not react directly with oxygen., 44. (c) : Air is liquefied by making use of the Joule, - Thomson effect (cooling by expansion of the gas)., , www.neetujee.com, , O, , 3, , 3, , 50. (c) : Peroxodisulphuric acid, H2S2O8 has, — O — O — linkage., , O OH, , Se— Se Te— Te, 172, 126, , O, , , , 48. (b) : NO2 is not used as a food preservative., , 49. (b) : Fe (SO )  Fe O + 3SO, , E in H2E increases, the bond H–E becomes weaker and, thus, breaks on heating. Therefore, the correct order of, thermal stability is H2Po < H2Te < H2Se < H2S < H2O., , B.E., O—O S—S, (kJ mol–1) : 142, 226, , , , , 47. (b) : Alkaline pyrogallol absorbs O2 and oil of, cinnamon absorbs O3., , O, S, , 40. (a) : Bond energy of S — S is exceptionally high due, to its catenation tendency., , O, , , , 38. (d) : The thermal stability of hydrides decreases, from H2O to H2Po. This is because as the size of atom, , 39. (a) : As the atomic size increases down the group,, the bond length increases and the bond strength, decreases and the cleavage of E–H bond becomes easier, thus, more will be the acidity. Thus, the correct order is :, H2S < H2Se < H2Te., , O, , , , , , O, , P, HO, , 116.8°, , O, , O, S, O, , O, , O, , OH, , 51. (a), 52. (a) :, , O, , O, , –, , O, –, , O S O S O, O, , O, , (S2O7, , 2–, , O, , –, , O S O O, O, , S O–, O, , 2–, , ), , (S2O8 ), , 53. (c) : Pyrosulphuric acid or oleum (+6) is H S O, 227, , which is obtained by dissolving SO3 and is called fuming, sulphuric acid., O, , O, , HO — S — O — S — OH, O, , O, , 54. (d), 55. (c) : All halogens show both positive and negative, oxidation states while fluorine shows only negative, oxidation state except +1 in HOF., , www.mediit.in

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52, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 63. (c) : Since chlorine is stronger oxidising agent than, bromine, therefore it will displace bromine from an, aqueous solution containing bromide ions., Cl2 + 2Br–  2Cl– + Br2, 64. (c) : Outer electronic configuration of Cl, = 3s23p 23p 23p 1, Outer electronic configuration of Cl–, = 3s23p 23p 23p 2, i.e., 4 lone pair of electrons, x, , 59. (a, d) : In case of diatomic molecules (X2) of halogens, the bond dissociation energy decreases in the order :, Cl2 > Br2 > F2 > I2. This is due to relatively large, electron-electron repulsion among the lone pairs is F2, than in case of Cl2., The oxidising power, electronegativity and reactivity, decrease in the order : F2 > Cl2 > Br2 > I2, Electron gain enthalpy of halogens follows the given, order :, Cl2 > F2 > Br2 > I2, The low value of electron gain enthalpy of fluorine is, probably due to small size of fluorine atom., 60. (a) : X – X bond F – F Cl – Cl Br – Br I – I, Bond dissociation, energy (kcal/mol), , 38, , 57, , 45.5, , 35.6, , The lower value of bond dissociation energy of fluorine, is due to the high inter-electronic repulsions between, non-bonding electrons in the 2p-orbitals of fluorine. As, a result F – F bond is weaker in comparison to Cl – Cl, and Br – Br bonds., 61. (a) : Due to more repulsion in between nonbonding electron pairs (2p) of two fluorines (due to small, size of F-atom) in comparison to non-bonding electron, pairs (3p) in chlorine, the bond energy of F 2 is less than, Cl2., B.E. (F2) = 158.8 kJ/mole and, B.E. (Cl2) = 242.6 kJ/mole, 62. (d) : In general, the electron affinity decreases from, top to bottom in a group. But in group 17, fluorine has, lower electron affinity as compared to chlorine due to, very small size of fluorine atom. As a result, there are, strong interelectronic repulsions in the relatively small 2s, orbitals of fluorine and thus, the incoming electron does, not experience much attraction., , www.neetujee.com, , y, , y, , z, , z, , 65. (a), –, , 0, , 66. (b) : 3Cl +(hot6NaOH, and conc.), , +5, , 5NaCl + NaClO + 3H2O, , This is an example of disproportionation reaction and, oxidation state of chlorine changes from 0 to –1 and +5., 67. (a) : MnO2 + 4HCl  MnCl2 + 2H2O + Cl2, 2KMnO4 + 16HCl  2KCl + 2MnCl2 + 8H2O + 5Cl2, 68. (a) : Chlorine is obtained by the electrolysis of brine, (concentrated NaCl solution). Chlorine is liberated at, anode., 69. (c) : Ca(OH)2 + Cl2  CaOCl2 + H2O, 70. (b) : Bromide in the mother liquor (containing, MgBr2) is oxidised to Br2 by passing Cl2 which is a, stronger oxidising agent., 2Br– + Cl2  Br2 + 2Cl–, 71. (d) : Bleaching action of chlorine is due to oxidation, in presence of moisture. Bleaching effect is permanent., H2O + Cl2  2HCl + [O], Colouring matter + [O]  Colourless matter, 72. (a) : CaOCl2 + 2HCl  CaCl2 + H2O + Cl2, 73. (d) : The acidic character of the oxoacids increases, with increase in oxidation number of the halogen atom, i.e., HClO < HClO2 < HClO3 < HClO4., This can be explained on the basis of relative stability, of the anions left after removal of a proton. Since, the stability of the anion decreases in the order :, ClO4– > ClO3– > ClO–2 > ClO–, acid strength also decreases, in the same order., 74. (c) : OF2 (oxygen difluoride) is a fluoride of oxygen, because fluorine is more electronegative than oxygen., 75. (d) :, .. –, .., .., Cl, O, Cl, O 110° O, Cl 111° Cl, O 118° O, –, .., , due to high electronegativity of F. Hence, the boiling, point of HF is abnormally high. Boiling points of other, hydrogen halides gradually increase from HCl to HI due, to increase in size of halogen atoms from Cl to I which, further increase the magnitude of van der Waals’ forces., 58. (d) : Standard reduction potentials of halogens are, positive and decrease from fluorine to iodine. So, F 2 is, the strongest oxidising agent., , x, , .., , 56. (d) : The order of bond dissociation enthalpy is :, Cl2 > Br2 > F2 > I2, B.E. (in kJ mol–1) 242.6 192.8 158.8 151.1, A reason for this anomaly is the relatively large electronelectron repulsion among the lone pairs in F2 molecule, where they are much closer to each other than in case, of Cl2., 57. (a) : HF forms strong intermolecular H-bonding, , Cl2 O, ClO2, ClO2, (2 lone pairs), (2 lone pairs), (1 lone pair), lower bond angle High electronegativity, Highest bond angle, of O than Cl so,, due to lone pair due to lower lone pair bond pair is more, lone pair, bond pair repulsion, closer to O atom so higher, lp - bp repulsion, , www.mediit.in

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The p-Block Elements, , 53, , 76. (d) :, , F, , O, , F, , F, , F, , Xe, F, , F, , XeOF4, Square pyramidal, sp3d2, , 77. (a), 78. (c) :, , 79. (a) :, , F, Xe, , O, F, , F, XeF6, Distorted octahedral, sp3d3, , Xe, O, , Xe, , F, , F, XeF4, Square planar, sp3d2, , O, , F, F, , F, , XeO3, Pyramidal, sp3, , O, , Tetrahedral (sp3 ), , O, O, , O, , 80. (a) : In XeF4 the ‘Xe’ atom is sp3d2 hybridised,, which contains two lone pair orbitals and four bond pair, orbitals. Therefore, the shape of XeF4 molecule is square, planar, with one lone pair orbital over and other below, the plane., , , www.neetujee.com, , www.mediit.in

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, , , , , , , CHAPTER, , 8, , The d - and f - Block, Elements, , 8.2 Electronic Configurations of the d-Block, Elements, , 1., , Sc (Z = 21) is a transition element but Zn (Z = 30) is, not because, (a) both Sc3+ and Zn2+ ions are colourless and form, white compounds, (b) in case of Sc, 3d orbitals are partially filled but in, Zn these are filled, (c) last electron is assumed to be added to 4s level, in case of Zn, (d) both Sc and Zn do not exhibit variable oxidation, states, (Karnataka NEET 2013), , 2., , Which of the following ions has electronic, configuration [Ar]3d6?, (a) Ni3+, (b) Mn3+ (c) Fe3+, (d) Co3+, (At. nos. Mn = 25, Fe = 26, Co = 27, Ni = 28) (2010), , 3., , Among the following series of transition metal ions,, the one where all metal ions have 3d2 electronic, configuration is, [At. nos. Ti = 22, V = 23, Cr = 24, Mn = 25], (a) Ti3+, V2+, Cr3+, Mn4+, (b) Ti+, V4+, Cr6+, Mn7+, (c) Ti4+, V3+, Cr2+, Mn3+, (d) Ti2+, V3+, Cr4+, Mn5+, (2004), , 4., , 5., , 6., , 7., , Which of the following configuration is correct for, iron?, (a) 1s22s22p63s23p64s23d7 (b) 1s22s22p63s23p64s23d5, (c) 1s22s22p63s23p63d5 (d) 1s22s22p63s23p64s23d6, (1999), Which of the following has more unpaird, d-electrons?, (a) N3+, (b) Fe2+, (c) Zn+, (d) Cu+, (1999), The electronic configuration of transition elements, is exhibited by, (a) ns1, (b) ns2np5, 2, 1–10, (c) ns (n – 1)d, (d) ns2 (n – 1)d10 (1996), , www.neetujee.com, , The electronic configurations of four elements are, given below. Which element does not belong to the, same family as others?, 10 2, (a) [Xe]4f145d106s2, (b) [Kr]4d 5s, (c) [Ne]3s23p5, (d) [Ar]3d104s2, (1989), , 8.3 General Properties of the Transition, Elements (d-Block), 8., , Identify the incorrect statement., (a) Cr2+ (d4) is a stronger reducing agent than, Fe2+ (d6) in water., (b) The transition metals and their compounds are, known for their catalytic activity due to their, ability to adopt multiple oxidation states and to, form complexes., (c) Interstitial compounds are those that are formed, when small atoms like H, C or N are trapped, inside the crystal lattices of metals., (d) The oxidation states of chromium in CrO4 2– and, Cr2O72– are not the same., (NEET 2020), , 9., , The calculated spin only magnetic moment of Cr 2+, ion is, (a) 3.87 BM, (b) 4.90 BM, (c) 5.92 BM, (d) 2.84 BM, (NEET 2020), , 10. Match the metal ions given in Column-I, with the spin magnetic moments of the ions given in, Column-II and assign the correct code :, Column-I, Column-II, A. Co3+, , (i), , 8 B.M., , B. Cr3+, , (ii), , 35 B.M., , C. Fe3+, , (iii) 3 B.M., , D. Ni2+, , (iv), , 24 B.M., , (v), , 15 B.M., , www.mediit.in

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The d - and f - Block Elements, A, (a) (iv), (b) (i), (c) (iv), (d) (iii), , 11., , 12., , 13., , 14., , 15., , B, (v), (ii), (i), (v), , C, (ii), (iii), (ii), (i), , 55, , D, (i), (iv), (iii), (ii), , 18. For the four successive transition elements (Cr, Mn,, Fe and Co), the stability of +2 oxidation state will be, there in which of the following order?, (a) Mn > Fe > Cr > Co (b) Fe > Mn > Co > Cr, (NEET 2018), (c) Co > Mn > Fe > Cr (d) Cr > Mn > Co > Fe, (At. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27) (2011), Magnetic moment 2.84 B.M. is given by, (At. nos. Ni = 28, Ti = 22, Cr = 24, Co = 27), 19. Which of the following ions will exhibit colour in, (a) Cr 2+ (b) Co2+, (c) Ni2+, (d) Ti3+, aqueous solutions?, (2015, Cancelled), (a) La3+ (Z = 57), (b) Ti3+ (Z = 22), Which of the following processes does not involve, 3+, (c) Lu (Z = 71), (d) Sc3+ (Z = 21) (2010), oxidation of iron?, 20. Which of the following pairs has the same size?, (a) Formation of Fe(CO)5 from Fe., (a) Fe2+, Ni2+, (b) Zr4+, Ti4+, (b) Liberation of H2 from steam by iron at high, (c) Zr4+, Hf4+, (d) Zn2+, Hf4+, (2010), temperature., (c) Rusting of iron sheets., 21. Which one of the elements with the following, (d) Decolourisation of blue CuSO 4 solution by, outer orbital configurations may exhibit the largest, iron., (2015, Cancelled), number of oxidation states?, Which of the following statements about the, (a) 3d54s1, (b) 3d54s2, 2 2, interstitial compounds is incorrect?, (c) 3d 4s, (d) 3d34s2, (2009), (a) They are much harder than the pure metal., 22. The correct order of decreasing second ionisation, (b) They have higher melting points than the pure, enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is, metal., (a) Mn > Cr > Ti > V (b) Ti > V > Cr > Mn, (c) They retain metallic conductivity., (c) Cr > Mn > V > Ti (d) V > Mn > Cr > Ti, (d) They are chemically reactive., (NEET 2013), (2008), Identify the alloy containing a non-metal as a, 23. In which of the following pairs are both the ions, constituent in it., coloured in aqueous solution?, (a) Invar, (b) Steel, (At. no. : Sc = 21, Ti = 22, Ni = 28, Cu = 29,Co = 27), (c) Bell metal, (d) Bronze, (2012), (a) Ni2+, Cu+, (b) Ni2+, Ti3+, The catalytic activity of transition metals and their, 3+, 3+, (c) Sc , Ti, (d) Sc3+, Co2+, (2006), compounds is ascribed mainly to, , (a) their magnetic behaviour, (b) their unfilled d-orbitals, (c) their ability to adopt variable oxidation states, (d) their chemical reactivity., (Mains 2012), 16. Which one of the following does not correctly, represent the correct order of the property indicated, against it?, (a) Ti < V < Cr < Mn; increasing number of, oxidation states, (b) Ti3+ < V3+ < Cr3+ < Mn3+ : increasing magnetic, moment, (c) Ti < V < Cr < Mn : increasing melting points, (d) Ti < V < Mn < Cr : increasing 2nd ionization, enthalpy, (Mains 2012), 17. Four successive members of the first series of the, transition metals are listed below. For which one of, them the standard potential (EM2+/M) value has a, positive sign?, (a) Co (Z = 27), (b) Ni (Z = 28), (c) Cu (Z = 29), (d) Fe (Z = 26), (Mains 2012), , www.neetujee.com, , 24. Four successive members of the first row transition, elements are listed below with their atomic numbers., Which one of them is expected to have the highest, third ionisation enthalpy?, (a) Vanadium (Z = 23), (b) Chromium (Z = 24), (c) Manganese (Z = 25), (d) Iron (Z = 26), (2005), 25. The aqueous solution containing which one of the, following ions will be colourless?, (Atomic number : Sc = 21, Fe = 26,Ti = 22, Mn = 25), (a) Sc3+, (b) Fe2+, (c) Ti3+, (d) Mn2+, (2005), 26. Which one of the following characteristics of the, transition metals is associated with their catalytic, activity?, (a) High enthalpy of atomization, (b) Paramagnetic behaviour, (c) Colour of hydrated ions, (d) Variable oxidation states, (2003), , www.mediit.in

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56, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 27. The basic character of the transition metal, monoxides follows the order, (Atomic no’s. Ti = 22, V = 23, Cr = 24, Fe = 26), (a) VO > CrO > TiO > FeO, (b) CrO > VO > FeO > TiO, (c) TiO > FeO > VO > CrO, (d) TiO > VO > CrO > FeO, (2003), 28. Which of the following shows maximum number of, oxidation states?, (a) Cr, (c) Mn, 29. Which ion is colourless?, (a) Cr4+, (c) Ti3+, , (b) Fe, (d) V (2002, 2000, 1994), (b) Sc3+, (d) V3+, , (2000), , 32. Which one of the following ionic species will impart, colour to an aqueous solution?, (a) Zn2+, (b) Cu+, 4+, (c) Ti, (d) Cr3+, (1998), 4, , 33. A transition element X has a configuration [Ar]3d, in its +3 oxidation state. Its atomic number is, (a) 22, (b) 19, (c) 25, (d) 26, (1996), 34. Amongst TiF2–, CoF3– , Cu Cl and NiCl2–, which, 6, , 2, , 2, , 4, , are the colourless species? (Atomic number of Ti =, 22, Co = 27, Cu = 29, Ni = 28), (a) CoF 3– and NiCl 2– (b) TiF2– and Cu Cl, 6, 4, 6, 2 2, (c) Cu2Cl2 and NiCl42– (d) TiF 2– and CoF 3–, 6, , 35. The mercury is the only metal which is liquid at 0°C., This is due to its, (a) high vapour pressure, (b) weak metallic bond, (c) high ionization energy, (d) both (b) and (c)., (1995), , Some Important, Transition Elements, , Compounds, , of, , 36. The manganate and permanganate ions are, tetrahedral, due to, (a) the -bonding involves overlap of d-orbitals of, oxygen with d-orbitals of manganese, (b) the -bonding involves overlap of p-orbitals of, oxygen with d-orbitals of manganese, , www.neetujee.com, , 3, , 38. Which one of the following ions exhibits d-d, transition and paramagnetism as well?, (a) CrO42–, (b) Cr 2O 72–, –, (c) MnO 4, (d) MnO 2–, 4 (NEET 2018), 39. Name the gas that can readily decolourise acidified, KMnO4 solution., (a) SO2, (b) NO2, (c) P2O5, (d) CO2, (NEET 2017), 40. Which one of the following statements is correct, when SO2 is passed through acidified K2Cr2O7, solution?, (a) SO2 is reduced., (b) Green Cr2(SO4)3 is formed., (c) The solution turns blue., (d) The solution is decolourised., (NEET-I 2016), 41. Assuming complete ionisation, same moles of which, of the following compounds will require the least, amount of acidified KMnO4 for complete oxidation?, (a) FeSO3, (b) FeC2O4, (c) Fe(NO2)2, (d) FeSO4, (2015), 42. The reaction of aqueous KMnO4 with H2O2 in acidic, conditions gives, (a) Mn4+ and O, (b) Mn2+ and O, 2+, , 2, , (c) Mn and O3, , 4+, , 2, , (d) Mn and MnO2., , (2014), , 6, , (1995), , 8.4, , 4, , (Odisha NEET 2019), , 30. Bell metal is an alloy of, (a) Cu + Zn, (b) Cu + Sn, (c) Cu + Pb, (d) Cu + Ni, (1999), 31. In which of the following compounds transition, metal has zero oxidation state?, (a) NOClO4, (b) NH2NH2, (c) CrO5, (d) [Fe(CO)5], (1999), , 6, , (c) there is no -bonding, (d) the -bonding involves overlap of p-orbitals of, oxygen with p-orbitals of manganese., (NEET 2019), 37. When neutral or faintly alkaline KMnO 4 is treated, with potassium iodide, iodide ion is converted into, ‘X’. ‘X’ is, (a) I2, (b) IO –, (c) IO–, (d) IO–, , 43. Which of the statements is not true?, (a) On passing H2S through acidified K2Cr2O7, solution, a milky colour is observed., (b) Na2Cr2O7 is preferred over K2Cr2O7 in, volumetric analysis., (c) K2Cr2O7 solution in acidic medium is orange., (d) K2Cr2O7 solution becomes yellow on increasing, the pH beyond 7., (2012), 44. Acidified K2Cr 2O7 solution turns green when, Na2SO3 is added to it. This is due to the formation of, (a) Cr2(SO4)3, (b) CrO2–, 4, (c) Cr2(SO3)3, (d) CrSO4, (2011), 45. The number of moles of KMnO4 reduced by one, mole of KI in alkaline medium is, (a) one, (b) two, (c) five, (d) one fifth., (2005), , www.mediit.in

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The d - and f - Block Elements, , 57, , 46. K2Cr2O7 on heating with aqueous NaOH gives, (a) Cr O2–, (b) Cr(OH), 2, , 7, , (c) CrO 2–, 4, , (d) Cr(OH), , 2, , 3, , (1997), , 47. KMnO4 reacts with oxalic acid according to the, equation, 2MnO–4 + 5C 2O2–, + 16H+, 2Mn2+ + 10CO +2 8H O2, 4, Here 20 mL of 0.1 M KMnO4 is equivalent to, (a) 50 mL of 0.5 M C2H2O4, (b) 20 mL of 0.1 M C2H2O4, (c) 20 mL of 0.5 M C2H2O4, (d) 50 mL of 0.1 M C2H2O4, (1996), 48. The oxidation state of Cr in K2Cr 2O7 is, (a) +5, (b) +3, (c) +6, (d) +7, , (1988), , 8.5 The Lanthanoids, 49. Which one of the following statements related to, lanthanons is incorrect?, (a) Europium shows +2 oxidation state., (b) The basicity decreases as the ionic radius, decreases from Pr to Lu., (c) All the lanthanons are much more reactive than, aluminium., (d) Ce(+4) solutions are widely used as oxidizing, agent in volumetric analysis., (NEET-II 2016), 50. The electronic configurations of Eu (Atomic No., 63), Gd (Atomic No. 64) and Tb (Atomic No. 65) are, (a) [Xe]4f 65d16s2, [Xe]4f 75d16s2 and [Xe]4f 85d16s2, (b) [Xe]4f 7 6s2, [Xe]4f 7 5d1 6s2 and [Xe]4f 9 6s2, (c) [Xe]4f 7 6s2, [Xe]4f 8 6s2 and [Xe]4f 8 5d1 6s2, (d) [Xe]4f 6 5d1 6s2, [Xe]4f 7 5d1 6s2 and [Xe]4f 9 6s2, (NEET-I 2016), 51. Gadolinium belongs to 4f series. Its atomic number, is 64. Which of the following is the correct electronic, configuration of gadolinium?, (a) [Xe] 4f 9 5s1, (b) [Xe] 4f 75d16s2, 6, 2 2, (c) [Xe] 4f 5d 6s, (d) [Xe] 4f 86d2, (2015, 1997), 52. Because of lanthanoid contraction, which of the, following pairs of elements have nearly same atomic, radii? (Numbers in the parenthesis are atomic, numbers), (a) Zr(40) and Hf(72) (b) Zr(40) and Ta(73), (c) Ti(22) and Zr(40) (d) Zr(40) and Nb(41), (2015, Cancelled), 53. Reason of lanthanoid contraction is, (a) negligible screening effect of ‘f ’-orbitals, (b) increasing nuclear charge, (c) decreasing nuclear charge, (d) decreasing screening effect., (2014), , www.neetujee.com, , 54. Which of the following lanthanoid ions is, diamagnetic?, (At. nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70), (a) Eu2+ (b) Yb2+, (c) Ce2+, (d) Sm2+, (NEET 2013), 55. Which of the following oxidation states is the most, common among the lanthanoids?, (a) 4, (b) 2, (c) 5, , (d) 3, (Mains 2010), , 56. Identify the incorrect statement among the following :, (a) Lanthanoid contraction is the accumulation of, successive shrinkages., (b) As a result of lanthanoid contraction, the, properties of 4d series of the transition elements, have no similarities with the 5d series of, elements., (c) Shielding power of 4f electrons is quite weak., (d) There is a decrease in the radii of the atoms or, ions as one proceeds from La to Lu. (2007), 57. Lanthanoids are, (a) 14 elements in the sixth period (atomic no. 90 to, 103) that are filling 4f sublevel, (b) 14 elements in the seventh period (atomic, number = 90 to 103) that are filling 5f sublevel, (c) 14 elements in the sixth period (atomic number, = 58 to 71) that are filling the 4f sublevel, (d) 14 elements in the seventh period (atomic, number = 58 to 71) that are filling 4f sublevel., (2004), 58. The correct order of ionic radii of Y3+, La3+, Eu3+ and, Lu3+ is (At. nos. Y = 39, La = 57, Eu = 63, Lu = 71), (a) Y3+ < La3+ < Eu3+ < Lu3+, (b) Y3+ < Lu3+ < Eu3+ < La3+, (c) Lu3+ < Eu3+ < La3+ < Y3+, (d) La3+ < Eu3+ < Lu3+ < Y3+, (2003), 59. General electronic configuration of lanthanides is, (a) (n – 2) f 1 – 14 (n – 1) s2p6d0 – 1 ns2, (b) (n – 2) f10 – 14 (n – 1) d0 – 1 ns2, (c) (n – 2) f 0 – 14 (n – 1) d10 ns2, (d) (n – 2) d0 – 1 (n – 1) f1 – 14 ns2, (2002), 60. Which of the following statement is not correct?, (a) La(OH)3 is less basic than Lu(OH)3., (b) In lanthanide series ionic radius of Ln+3 ion, decreases., (c) La is actually an element of transition series, rather lanthanides., (d) Atomic radius of Zn and Hf are same because of, lanthanide contraction., (2001), , www.mediit.in

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58, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 61. The lanthanide contraction is responsible for the, fact that, (a) Zr and Hf have about the same radius, (b) Zr and Zn have the same oxidation state, (c) Zr and Y have about the same radius, (d) Zr and Nb have similar oxidation state. (1997), 62. Which of the following statements concerning, lanthanide elements is false?, (a) All lanthanides are highly dense metals., (b) More characteristic oxidation state of lanthanide, elements is +3., (c) Lanthanides are separated from one another by, ion exchange method., (d) Ionic radii of trivalent lanthanides steadily, increases with increase in the atomic number., (1994), , 8.6 The Actinoids, 63. The reason for greater range of oxidation states in, actinoids is attributed to, (a) actinoid contraction, (b) 5f, 6d and 7s levels having comparable energies, (c) 4f and 5d levels being close in energies, (d) the radioactive nature of actinoids., (NEET 2017), 64. Which of the following exhibits only +3 oxidation, state?, (a) U, (b) Th, (c) Ac, (d) Pa, (Mains 2012), 65. More number of oxidation states are exhibited by, the actinoids than by the lanthanoids. The main, reason for this is, (a) more active nature of the actinoids, (b) more energy difference between 5f and 6d, orbitals than that between 4f and 5d orbitals, (c) lesser energy difference between 5f and 6d, orbitals than that between 4f and 5d orbitals, (d) greater metallic character of the lanthanoids, than that of the corresponding actinoids., (2006, 2005), , 66. Which one of the following elements shows, maximum number of different oxidation states in, its compounds?, (a) Gd, (b) La, (c) Eu, (d) Am, (1998), , 8.7 Some Applications of d- and f-Block, Elements, 67. Match the catalyst with the process :, Catalyst, Process, (i) V2O5, (p) The oxidation of ethyne to, ethanal, (ii) TiCl4 +Al(CH3)3 (q) Polymerisation of alkynes, (iii) PdCl2, (r) Oxidation of SO2 in the, manufacture of H2SO4, (iv) Nickel, (s) Polymerisation, complexes, of ethylene, Which of the following is the correct option?, (a) (i)-(r), (ii)-(s), (iii)-(p), (iv)-(q), (b) (i)-(p), (ii)-(q), (iii)-(r), (iv)-(s), (c) (i)-(p), (ii)-(r), (iii)-(q), (iv)-(s), (d) (i)-(r), (ii)-(p), (iii)-(s), (iv)-(q), (Odisha NEET 2019), 68. HgCl2 and I2– both when dissolved in water, containing –I ions, the pair of species, formed is, (a) HgI2, I, (b) HgI42–, I3–, –, (c) Hg2I2, I, (d) HgI , I–, 23, , (NEET 2017), 69. Which of the following elements is responsible for, oxidation of water to O2 in biological processes?, (a) Cu, (b) Mo, (c) Fe, (d) Mn, (1997), 70. When calomel reacts with NH4OH, we get, (a) Hg2O, (b) HgO, (c) HgNH2Cl, (d) NH2–Hg–Hg–Cl, (1996), 71. Photographic films and plates have an essential, ingredient of, (a) silver nitrate, (b) silver bromide, (c) sodium chloride, (d) oleic acid., (1989), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., 61., 71., , (b), (c), (b), (d), (d), (b), (a), (b), , www.neetujee.com, , 2., 12., 22., 32., 42., 52., 62., , (d), (a), (c), (d), (b), (a), (d), , 3., 13., 23., 33., 43., 53., 63., , (d), (d), (b), (c), (b), (a), (b), , 4., 14., 24., 34., 44., 54., 64., , (d), (b), (c), (b), (a), (b), (c), , 5., 15., 25., 35., 45., 55., 65., , (b), (c), (a), (d), (b), (d), (c), , 6., 16., 26., 36., 46., 56., 66., , (c), (c), (d), (b), (c), (b), (d), , 7., 17., 27., 37., 47., 57., 67., , (c), (c), (d), (c), (d), (c), (a), , 8., 18., 28., 38., 48., 58., 68., , (d), (a), (c), (d), (c), (b), (b), , 9., 19., 29., 39., 49., 59., 69., , (b), (b), (b), (a), (c), (a), (c), , 10., 20., 30., 40., 50., 60., 70., , (a), (c), (b), (b), (b), (a), (c), , www.mediit.in

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The d - and f - Block Elements, , 59, , Hints & Explanations, 1. (b) : Sc (Z = 21) has incompletely filled 3d-orbitals, in its ground state (3d1), it is considered as a transition, element but Zn (Z = 30) has completely filled d-orbitals, (3d10) in its ground state and its common oxidation state, (+2), thus, it is not considered as a transition element., , 14. (b) : Invar  Ni(metal) + Fe(metal), Steel  C(non-metal) + Fe(metal), Bell  Cu(metal) + Sn(metal) + Fe(metal), Bronze  Cu(metal) + Sn(metal), , 2. (d) : The electronic configuration of the given ions, is :, Ni3+ : [Ar]3d74s0, Mn3+ : [Ar]3d44s0, Fe3+ : [Ar]3d54s0, Co3+ : [Ar]3d64s0, Thus, Co3+ is the ion with the desired configuration., , 16. (c) : Element :, Ti < V < Cr < Mn, No. of oxidation states : +3 +4 +5 +6, Hence, given order is correct., , (d) : 22Ti = 3d2 4s2 ; Ti2+ = 3d2, 3, 2, 3+, = 3d2 2, 23V = 3d 44s 2; V 4+, 24Cr = 3d 4s ; Cr = 3d, Mn = 3d5 4s2 ; Mn5+ = 3d2, , 3., , 25, , 4., , (d), , 5. (b), , 6. (c) : General electronic configuration of transition, elements is ns2 (n – 1)d1 – 10., 7. (c) : [Ne]3s23p5 is the electronic configuration of a, p-block element whereas other configurations are those, of d-block elements., 8. (d) : The oxidation states of Cr in CrO 2– and Cr O2–, 4, 2 7, is same i.e., +6., 9. (b) : Cr : 3d5 4s1, Cr2+ : 3d4 has four unpaired, electrons.,  n(n  2)  4(4  2)  24   , 10. (a) : Co3+ : [Ar]3d6, unpaired e–(n) = 4, 4(4  2)  24 B. M., Cr : [Ar]3d , unpaired e (n) = 3, Spin magnetic moment () =, 3+, , 3, , –, , Spin magnetic moment () = 3(3  2)  15 B. M., Fe3+ : [Ar]3d5, unpaired e–(n) = 5, Spin magnetic moment ()  5(5  2)  35 B. M., Ni2+ : [Ar]3d8, unpaired e–(n) = 2, Spin magnetic moment ()  2(2  2)  8 B. M., 11. (c) : Magnetic moment () = n(n  2), 2.84 B.M. corresponds to 2 unpaired electrons., Cr2+ – 3d4, 4 unpaired electrons, Co2+ – 3d7, 3 unpaired electrons, Ni2+ – 3d8, 2 unpaired electrons, Ti3+ – 3d1, 1 unpaired electron, 12. (a) : Oxidation number of Fe in Fe(CO)5 is zero., 13. (d) : Interstitial compounds are generally chemically, inert., , www.neetujee.com, , 15. (c), , Magnetic moment ()  n(n  2) B.M., For Ti3+ n = 1,  = 1(1  2)  3 B.M., For V3+ n = 2,  = 2(2  2)  8 B.M., For Cr3+ n = 3,  = 3(3  2)  15 B.M., For Mn3+ n = 4,  =, , 4(4  2)  24 B.M., , Thus, magnetic moment order : Ti3+ < V3+ < Cr3+ < Mn3+, Melting point order : Mn < Ti < Cr < V, 1245°C 1668°C 1875°C 1900°C, , 2nd ionisation enthalpy order, Ti < V < Mn < Cr, (in kJ/mol) :, 1309 1414, 1509, 1592, 17., : Element – 0.28, Co –0.25, Ni +0.34, Cu –0.44Fe, E° 2+(c) (V), M /M, , 18. (a) : Spin correlation and exchange energy gives, an electronic configuration a special stability which is, greatest for half-filled electronic configurations., Mn2+ (d5) gets stabilisation due to half-filled, configuration., In Fe2+ (d6) the placing of one extra electron in a subshell, destabilises. Placing of 2 electrons in Co2+ (d7) destabilises, it more. Cr2+ (d4) has one vacant subshell. Fe2+ gets more, stabilisation compared to Cr2+ through exchange energy., So, the order is as follows : Mn > Fe > Cr > Co., 19. (b) : Ions which have unpaired electrons exhibit, colour in aqueous solution. Ti3+ has an outer electronic, configuration of 4s03d1, i.e., 1 unpaired electron. Thus,, its solution will be coloured. Others are colourless due to, empty or completely filled outermost orbitals., 20. (c) : Hf 4+ and Zr 4+ belong to group IVB. But, Hf 4+, has same size as Zr4+ due to the addition of 14 lanthanide, elements before it in which electrons are added into the, f-subshell which poorly shield the outer electrons and, contraction in size occurs., 21. (b) : Greater the number of valence electrons, more, will be the number of oxidation states exhibited by the, element., 3d54s1, can show a maximum of 6 oxidation states., , www.mediit.in

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The d - and f - Block Elements, , 61, , 40. (b) : K2Cr2O7 + H2SO4 + 3SO2  K2SO4, + Cr (SO ) + H O, 2, , 4 3, , 2, , (Green), , 41. (d) : KMnO4 (Mn7+) changes to Mn2+ i.e., number, of electrons involved per mole of KMnO4 is 5., (a) For FeSO3,, Fe2+ Fe3+ (No. of e–s involved = 1), SO32– SO2– 4 (No. of e–s involved = 2), Total number of e–s involved = 1 + 2 = 3, (b) For FeC2O4,, Fe2+, Fe3+, (No. of e–s involved = 1), 2–, C2O 4 2CO (No. 2of e–s involved = 2), Total number of e–s involved = 1 + 2 = 3, (c) For Fe(NO2)2,, Fe2+, Fe3+, (No. of e–s involved = 1), –, –, 2NO2, 2NO 3(No. of e–s involved = 4), Total number of e–s involved = 1 + 4 = 5, (d) For FeSO4,, Fe2+, Fe3+, (No. of e–s involved = 1), –, Total number of e s involved = 1, As FeSO4 requires least number of electrons thus, it will, require least amount of KMnO4., 42. (b) : Hydrogen peroxide is oxidised to H 2O and O2., 2KMnO4 + 3H2SO4 + 5H2O2, K2SO4 + 2MnSO4, + 8H2O + 5O2, or, 2MnO – + 5H O + 6H+, 2Mn2+ + 8H O + 5O, 4, , 2, , 2, , 2, , 2, , 43. (b) : Potassium dichromate is preferred over sodium, dichromate in volumetric analysis, primarily because, the latter is hygroscopic nature and therefore, accurate, weighing is not possible in normal atmosphere., 44. (a) : K2Cr2O7 + 4H2SO4, , K2SO4 + Cr2(SO4)3, + 4H2O + 3[O], Na2SO4] × 3, , [Na2SO3 + [O], , K2Cr2O7 + 3Na2SO3 + 4H2SO4, or, , Cr O2– + 3SO2– + 8H+, 2, , 7, , 3, , 3Na2SO4 + K2SO4, + Cr2(SO4)3 + 4H2O, 3SO2– + 2Cr3+ + 4H O, 4, , 2, , 45. (b) : In alkaline medium :, 2KMnO4 + H2O  2KOH + 2MnO2 + 3[O], KI + 3[O]  KIO3, 2KMnO + KI + H O  2KOH + 2MnO + KIO, 4, , 2, , 2, , 3, , 46. (c) : K2Cr2O7 + 2NaOH  K2CrO4 + Na2CrO4, + H2O, –, 2–, or Cr 2O2–, +, 2OH, , 2CrO, +, H, O, 2, 7, 4, 47. (d) : 2MnO– + 5C O2– + 16H+  2Mn2+ + 10CO, 4, , , , 2 4, , www.neetujee.com, , 48. (c) : Let, oxidation state of Cr in K2Cr2O7 is x. Then,, 2 + 2x – 14 = 0,  2x = 12,  x = +6, 49. (c) : The first few members of the lanthanoid series, are quite reactive, almost like calcium. However, with, increasing atomic number, their behaviour becomes, similar to that of aluminium., 50. (b), 51. (b), 52. (a) : Zr and Hf have nearly same radii due to, lanthanoid contraction., 53. (a) : Due to poor shielding effect of 4f-orbitals,, nucleus will exert a strong attraction and size of atom or, ion goes on decreasing as move in the series with increase, in atomic number., 54. (b) : Sm2+ (Z = 62) : [Xe]4f 6, Eu2+ (Z = 63) : [Xe]4f 7, Yb2+ (Z = 70) : [Xe]4f 14, Ce2+ (Z = 58) : [Xe]4f 2, 2+, Only Yb is diamagnetic., 55. (d) : The common stable oxidation state of all the, lanthanoids is +3. The oxidation state of +2 and +4 are, also exhibited by some of the elements. These oxidation, states are only stable in those cases where stable 4f 0, 4f 7, or 4f 14 configurations are achieved., 56. (b) : In each vertical column of transition elements,, the elements of second and third transition series, resemble each other more closely than the elements of, first and second transition series on account of lanthanide, contraction. Hence, the properties of elements of 4d series, of the transition elements resemble with the properties of, the elements of 5d series of the transition elements., 57. (c) : As sixth period can accommodate only 18, elements in the table, 14 members of 4f series (atomic, number 58 to 71) are separately accommodated in a, horizontal row below the periodic table. These are called, as lanthanides., 58. (b) : On going from La3+ to Lu3+, the ionic radius, shrinks from 1.15 Å to 0.93 Å (lanthanide contraction)., The radius of La3+ is also larger than that of Y3+ ion which, lies immediately above it in periodic table., 59. (a) : The general electronic structure of lanthanides, is, (n – 2)f 1 – 14 (n – 1)s2p6d0 –1ns2., 60. (a) : La(OH)3 is more basic than Lu(OH)3. In, , 2, , 2 moles of MnO–  5 moles of C O2–, 4, , 20 mL of 0.1 M KMnO4 = 2 mmol of KMnO4, 2–, Also, 50 mL of 0.1 M C2H2O4  5 mmol of C, 2 4O, Therefore, these are equivalent., , + 8H2O, , lanthanides, the basic character of hydroxides decreases, as the ionic radius decreases., , 2 4, , www.mediit.in

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62, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 61. (a) : Due to lanthanide contraction, the elements, of second and third transition series i.e., Zr and Hf, resemble more with each other than the elements of first, and second transition series., 62. (d) : Ionic radii of trivalent lanthanides decreases, with increase in atomic number., 63. (b) : Actinoids have a greater range of oxidation, states due to comparable energies of 5f, 6d and 7s, orbitals. Hence, all their electrons can take part in bond, formation., , 66. (d) : ‘La’ forms compounds in which its oxidation, no. is +3., ‘Eu’ and ‘Gd’ exhibit +2 as well as +3 oxidation, states and not higher than that, due to stable, (f 7) configuration. whereas ‘Am’ exhibits the oxidation, states +3, +4, +5, +6, etc. due to extremely large size and, low ionisation energy., 67. (a), 68. (b) : HgCl2( aq) + 4I– aq), I + I–, I– (, 2(s), , 64. (c) : U exhibits + 3, + 4, + 5, + 6, Th exhibits + 3, + 4 ; Ac exhibits + 3 only, Pa exhibits + 3, + 4, + 5, , (aq), , HgI 2– aq) + 2Cl– aq), 4 (, , (, , 3 (aq), , 69. (c), , 65. (c) : The 5f-orbitals extend into space beyond the, 6s and 6p-orbitals and participate in bonding. This is in, direct contrast to the lanthanides where the 4f-orbitals, are buried deep inside in the atom, totally shielded by, outer orbitals and thus, unable to take part in bonding., , 70. (c) : When calomel reacts with NH4OH, it turns, black due to the formation of a mixture of mercury and, ammonium basic mercury (II) chloride., Hg2Cl2 + 2 NH4OH  NH4Cl + 2H2O + Hg + HgNH2Cl, (Calomel), , 71. (b) : AgBr is highly photosensitive and is used as an, ingredient for photographic films and plates., , , , www.neetujee.com, , www.mediit.in

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, , , , , , , CHAPTER, , Coordination, Compounds, , 9, , 9.1, , 9.2, , Werner’s Theory of Coordination, Compounds, , 1. The correct order of the stoichiometries of AgCl, formed when AgNO3 in excess is treated with the, complexes, respectively: CoCl, is 3.6NH3, CoCl3.5NH3, CoCl3.4NH3, , 6., , 2., , 3., , –, , –, , –, , –, , –, , –, , (d) CN > Br > C6H5 > NH3, (NEET 2017), , Cobalt(III) chloride forms several octahedral, complexes with ammonia. Which of the following, will not give test for chloride ions with silver nitrate, at 25°C?, (a) CoCl3 · 5NH3, (b) CoCl3 · 6NH3, (c) CoCl3 · 3NH3, (d) CoCl3 · 4NH3, (2015, Cancelled), An excess of AgNO3 is added to 100 mL of a 0.01 M, solution of dichlorotetraaquachromium(III) chloride., The number of moles of AgCl precipitated would be, (a) 0.003, (b) 0.01, (c) 0.001, (d) 0.002 (NEET 2013), , 4., , Which of the following will exhibit maximum ionic, conductivity?, (a) K4[Fe(CN)6], (b) [Co(NH3)6]Cl3, (c) [Cu(NH3)4]Cl2 (d) [Ni(CO)4] (2001), , 5., , A coordination complex compound of cobalt has, the molecular formula containing five ammonia, molecules, one nitro group and two chlorine atoms, for one cobalt atom. One mole of this compound, produces three mole ions in an aqueous solution., On reacting this solution with excess of AgNO3, solution, we get two moles of AgCl precipitate. The, ionic formula for this complex would be, (a) [Co(NH 3) 5(NO2)]Cl2, (b) [Co(NH3)5Cl][Cl(NO2)], (c) [Co(NH 3) 4(NO2)Cl][(NH 3)Cl], (d) (Co(NH3)5][(NO2)2Cl2], (1998), , www.neetujee.com, , The correct increasing order of trans-effect of the, following species, is, –, (a) NH3 > CN – > Br– > C6 H, 5, (b) CN, > C6H5 > Br > NH3, (c) Br – > CN – > NH 3 > 6C5 H, , (a) 3AgCl, 1AgCl, 2AgCl, (b) 3AgCl, 2AgCl, 1AgCl, (c) 2AgCl, 3AgCl, 2AgCl, (d) 1AgCl, 3AgCl, 2AgCl, , Definitions of Some Important Terms, Pertaining to Coordination Compounds, , –, , (NEET-II 2016), , 7., , The sum of coordination number and oxidation, number of the metal M in the complex, [M(en)2(C2O4)]Cl (where en is ethylenediamine) is, (a) 6, (b) 7, (c) 8, (d) 9, (2015), , 8., , The anion of acetylacetone (acac) forms Co(acac) 3, chelate with Co3+. The rings of the chelate are, (a) five membered (b) four membered, (c) six membered, (d) three membered., (Karnataka NEET 2013), , 9., , Which of the following statements is true?, (a) Silicon exhibits 4 coordination number in its, compound., (b) Bond energy of F2 is less than Cl2., (c) Mn(III) oxidation state is more stable than, Mn(II) in aqueous state., (d) Elements of 15th gp shows only +3 and +5, oxidation states., (2002), , 10., , Coordination number of Ni in [Ni(C2O4)3]4– is, (a) 3, (b) 6, (c) 4, (d) 2, (2001), , 11., , The coordination number and oxidation state of Cr, in K3[Cr(C2O4)3] are respectively, (a) 3 and + 3, (b) 3 and 0, (c) 6 and + 3, (d) 4 and+ 2, (1995), , 12., , Which of the following ligands is expected to be, bidentate?, (a) CH3NH 2, (b) CH3 C N, (c) Br, (d) C2O 42–, (1994), , www.mediit.in

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64, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 9.3, 13., , 14., , 15., , 16., , 17., , 19., , of, , Coordination, , The name of complex ion, [Fe(CN)6]3– is, (a) hexacyanitoferrate(III) ion, (b) tricyanoferrate(III) ion, (c) hexacyanidoferrate(III) ion, (d) hexacyanoiron(III) ion., , (2015), , The correct IUPAC name for [CrF2(en)2]Cl is, (a) chlorodifluoridoethylenediaminechromium, (III) chloride, (b) difluoridobis(ethylene diamine)chromium (III), chloride, (c) difluorobis-(ethylene diamine)chromium (III), chloride, (d) chlorodifluoridobis(ethylene diamine), chromium (III)., (Karnataka NEET 2013), The hypothetical complex chlorodiaquatriammine, cobalt(III) chloride can be represented as, (a) [CoCl(NH 3) 3(H2O) 2]Cl2, (b) [Co(NH 3) 3(H2O)Cl3], (c) [Co(NH 2) 3(H2O) 2Cl], (d) [Co(NH3)3(H2O)3]Cl3, (2002), IUPAC name of [Pt(NH3)3(Br)(NO2)Cl]Cl is, (a) triamminebromochloronitroplatinum(IV), chloride, (b) triamminebromonitrochloroplatinum(IV), chloride, (c) triamminechlorobromonitroplatinum(IV), chloride, (d) triamminenitrochlorobromoplatinum(IV), chloride., (1998), The formula of dichlorobis(urea)copper(II) is, (a) [Cu {O C(NH2)2} Cl]Cl, (b) [CuCl2] {O C(NH2)2}, (c) [Cu {O C(NH2)2}Cl2, (d) [CuCl2 {O C(NH2)2}2], (1997), , 9.4, 18., , Nomenclature, Compounds, , 20., , The, complexes, [Co(NH3)6][Cr(CN)6], and, [Cr(NH3)6][Co(CN)6] are the examples of which, type of isomerism?, (a) Linkage isomerism, (b) Ionization isomerism, (c) Coordination isomerism, (d) Geometrical isomerism, (2011), , 21., , The complex, [Pt(py)(NH3)BrCl] will have how, many geometrical isomers?, (a) 3, (b) 4, (c) 0, (d) 2, (2011), , 22., , The existence of two different coloured complexes, with the composition of [Co(NH3)4Cl2]+ is due to, (a) linkage isomerism, (b) geometrical isomerism, (c) coordination isomerism, (d) ionization isomerism., (2010), , 23., , Which one of the following complexes is not, expected to exhibit isomerism?, (a) [Ni(NH 3) 4(H2O) 2] 2+ (b) [Pt(NH 3) 2Cl2], (c) [Ni(NH3)2Cl2], (d) [Ni(en)3]2+, (2010), , 24., , Which of the following does not show optical, isomerism?, (a) [Co(NH 3) 3Cl3] 0, (b) [Co(en)Cl2(NH3)2]+, (c) [Co(en)3]3+, (d) [Co(en)2Cl2]+(en = ethylenediamine) (2009), , 25., , Which of the following will give a pair of, enantiomorphs?, (a) [Cr(NH 3) 6][Co(CN) 6], (b) [Co(en)2Cl2]Cl, (c) [Pt(NH 3) 4][PtCl6], (d) [Co(NH 3) 4Cl2]NO2, (en = NH2CH2CH2NH2), , 26., , [Co(NH3)4(NO2)2]Cl exhibits, (a) linkage isomerism, geometrical isomerism and, optical isomerism, (b) linkage isomerism, ionization isomerism and, optical isomerism, (c) linkage isomerism, ionization isomerism and, geometrical isomerism, (d) ionization isomerism, geometrical isomerism, and optical isomerism., (2006), , 27., , Which one of the following is expected to exhibit, optical isomerism? (en = ethylenediamine), (a) cis-[Pt(NH3)2Cl2], (b) trans-[Pt(NH3)2Cl2], (c) cis-[Co(en)2Cl2]+, (d) trans-[Co(en)2Cl2]+, (2005), , Isomerism in Coordination Compounds, , The type of isomerism shown by the complex, [CoCl2(en)2] is, (a) geometrical isomerism, (b) coordination isomerism, (c) ionization isomerism, (d) linkage isomerism., (NEET 2018), Number of possible isomers for the complex, [Co(en)2Cl2]Cl will be (en = ethylenediamine), (a) 1, (b) 3, (c) 4, (d) 2, (2015), , www.neetujee.com, , (2007), , www.mediit.in

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65, , Coordination Compounds, , 28., , Which of the following coordination compounds, would exhibit optical isomerism?, (a) Pentaamminenitrocobalt(III) iodide, (b) Diamminedichloroplatinum(II), (c) trans-Dicyanobis(ethylenediamine), chromium(III) chloride, (d) tris-(Ethylenediamine)cobalt(III) bromide, (2004), , 29., , Which of the following will give maximum number, of isomers?, (a) [Co(NH 3) 4Cl2], (b) [Ni(en)(NH 3) 4] 2+, 2–, (c) [Ni(C2O 4)(en) 2], (d) [Cr(SCN) 2(NH3) 4] +, (2001), , 30., , Which complex compound will give four isomers?, (a) [Fe(en)3]Cl3, (b) [Co(en)2Cl2]Cl, (c) [Fe(PPh3)3NH3ClBr]Cl, (d) [Co(PPh3)3Cl]Cl3, (2000), , 31., , The total number of possible isomers for the complex compound, [CuII(NH 3) 4][PtIICl ]4 are, (a) 5, (b) 6, (c) 3, (d) 4, (1998), , 32. The number of geometrical isomers of the complex, [Co(NO2)3(NH3)3] is, (a) 4, (b) 0, (c) 2, (d) 3, (1997), 33. The number of, [Pt(NH3)2Cl2] is, (a) 3, (c) 1, , 9.5, , geometrical, (b) 4, (d) 2, , isomers, , for, , (1995), , Bonding in Coordination Compounds, , 34. Which of the following is the correct order, of increasing field strength of ligands to form, coordination compounds?, –, (a) SCN–– < F–– < C2O2–, 4 < CN2–, –, (a) SCN, < F < CN < C2O4, (b) F– < SCN– < C 2O2–4 < CN–, –, , 2–, , –, , –, , (c) CN < C2O4 < SCN < F, , (NEET 2020), , 35. What is the correct electronic configuration of the, central atom in K4[Fe(CN)6] based on crystal field, theory?, 4 2, (a) e4t 2, (b) t e, 2, , (c) t6 e0, 2g g, , 2g g, , (d) e3 t3, , (NEET 2019), , 2, , 36. Aluminium chloride in acidified aqueous solution, forms a complex ‘A’, in which hybridisation state of, Al is ‘B’. What are ‘A’ and ‘B’, respectively ?, (a) [Al(H2O)6]3+, sp3d2, (b) [Al(H2O)4]3+, sp3, , www.neetujee.com, , (c) [Al(H2O)4]3+, dsp2, (d) [Al(H2O)6]3+, d2sp3, , (Odisha NEET 2019), , 37. The crystal field stabilisation energy (CFSE) for, [CoCl6]4– is 18000 cm–1. The CFSE for [CoCl 4]2–, will be, (a) 6000 cm–1, (b) 16000 cm–1, –1, (c) 18000 cm, (d) 8000 cm–1, (Odisha NEET 2019), 38. The geometry and magnetic behaviour of the, complex [Ni(CO)4] are, (a) square planar geometry and diamagnetic, (b) tetrahedral geometry and diamagnetic, (c) square planar geometry and paramagnetic, (d) tetrahedral geometry and paramagnetic., (NEET 2018), 39. Correct increasing order for the wavelengths of, absorption in the visible region for the complexes of, Co3+ is, (a) [Co(H 2O) 6] 3+, [Co(en) 3] 3+, [Co(NH 3) 6] 3+, (b) [Co(H 2O) 6] 3+, [Co(NH 3)6] 3+, [Co(en) 3] 3+, (c) [Co(NH3)6]3+, [Co(en)3]3+, [Co(H2O)6]3+, (d) [Co(en) 3] 3+, [Co(NH 3) 6] 3+, [Co(H 2O) 6] 3+, (NEET 2017), 40. Pick out the correct statement with respect to, [Mn(CN) 6] 3–., (a) It is sp3d2 hybridised and tetrahedral., (b) It is d2sp3 hybridised and octahedral., (c) It is dsp2 hybridised and square planar., (d) It is sp3d2 hybridised and octahedral., (NEET 2017), 41. Jahn–Teller effect is not observed in high spin, complexes of, 8, (a) d7, (b) d, (c) d4, (d) d9, (NEET-II 2016), 42. The hybridization involved in complex [Ni(CN)4]2–, is (At.3No. Ni = 28), (a) sp, , (b) d2sp2, , (c) d2sp3, , (d) dsp2, , (2015), , 43. Among the following complexes the one which, shows zero crystal field stabilization energy (CFSE), is, (a) [Mn(H 2O) 6] 3+, (b) [Fe(H 2O) 6] 3+, 2+, , (c) [Co(H 2O) 6], , 3+, , (d) [Co(H 2O) 6], , (2014), , 44. A magnetic moment at 1.73 BM will be shown by, one among of the following, (a) TiCl4, (b) [CoCl6] 4–, (c) [Cu(NH3)4]2+, (d) [Ni(CN)4]2–, (NEET 2013), , www.mediit.in

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66, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 45. Crystal field splitting energy for high spin d 4, octahedral complex is, (a) – 1.2 o, (b) – 0.6 o, (c) – 0.8 o, (d) – 1.6 o, (Karnataka NEET 2013), 46. Which among the following is a paramagnetic, complex?, (a) [Co(NH 3) 6] 3+, (b) [Pt(en)Cl2], (c) [CoBr4]2–, (d) Mo(CO)6, (At. No. Mo = 42, Pt = 78), (Karnataka NEET 2013), 47. Which is diamagnetic?, (a) [CoF 6] 3–, (b) [Ni(CN) 4] 2–, 2–, (c) [NiCl4], (d) [Fe(CN)6]3–, (Karnataka NEET 2013), 48. Which one of the following is an outer orbital, complex and exhibits paramagnetic behaviour?, (a) [Ni(NH 3) 6] 2+, (b) [Zn(NH 3) 6] 2+, (d) [Co(NH 3) 6] 3+ (2012), (c) [Cr(NH 3) 6] 3+, 49. Low spin complex of d6-cation in an octahedral field, will have the following energy, 12, 12, (a), (b), o  3P, o  P, , 5, 5, 2, 2, (d), P, (c),   2P, 5 o, 5 o, (o = crystal field splitting energy in an octahedral, field, P = Electron pairing energy), (2012), 50. Red precipitate is obtained when ethanol solution of, dimethylglyoxime is added to ammoniacal Ni(II)., Which of the following statements is not true?, (a) Red complex has a square planar geometry., (b) Complex has symmetrical H-bonding., (c) Red complex has a tetrahedral geometry., (d) Dimethylglyoxime functions as bidentate, ligand., , (Mains 2012), 51. Of the following complex ions, which is diamagnetic, in nature?, (a) [NiCl4]2–, (b) [Ni(CN)4]2–, 2–, (c) [CuCl4], (d) [CoF6]3–, (2011), 52. The d-electron configurations of Cr2+, Mn2+,, Fe2+ and Co2+ are d4, d5, d6 and d7 respectively., Which one of the following will exhibit minimum, paramagnetic behaviour?, (a) [Mn(H 2O) 6] 2+, (b) [Fe(H 2O) 6] 2+, , www.neetujee.com, , (c) [Co(H2O) 6] 2+, (d) [Cr(H 2O) 6] 2+, (At. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27) (2011), 53. Which of the following complex compounds will, exhibit highest paramagnetic behaviour?, (a) [Ti(NH 3) 6] 3+, (b) [Cr(NH 3) 6] 3+, (c) [Co(NH 3) 6] 3+, (d) [Zn(NH 3) 6] 2+, (At. No. Ti = 22, Cr = 24, Co = 27, Zn = 30) (2011), 54. Which of the following complex ions is not expected, to absorb visible light?, (a) [Ni(CN) 4] 2–, (b) [Cr(NH 3) 6] 3+, 2+, (c) [Fe(H 2O) 6], (d) [Ni(H 2O) 6] 2+ (2010), 55. Crystal field stabilization energy for high spin d4, octahedral complex is, (a) – 1.8 o, (b) – 1.6 o + P, (c) – 1.2 o, (d) – 0.6 o, (2010), 56. Out of TiF2–, CoF 3–, Cu Cl and NiCl 2– (Z of, 6, , 6, , 2, , 2, , 4, , Ti = 22, Co = 27, Cu = 29, Ni = 28) the colourless, species are, (a) Cu2Cl3–2 and NiCl 2–42– (b) TiF2–2– 6and Cu Cl, 2 2, (c) CoF and NiCl (d) TiF and CoF 3–., 6, , 4, , 6, , 6, , (2009), 57. Which of the following complex ions is expected to, absorb visible light?, (a) [Ti(en) 2(NH3) 2] 4+ (b) [Cr(NH 3) 6] 3+, (c) [Zn(NH 3) 6] 2+, (d) [Sc(H 2O) 3(NH3) 3] 3+, [At. nos. Zn = 30, Sc = 21, Ti = 22, Cr = 24] (2009), 58. Which of the following complexes exhibits the, highest paramagnetic behaviour?, (a) [Co(ox)2(OH)2]–, (b) [Ti(NH3)6]3+, (c) [V(gly)2(OH)2(NH3)2]+, (d) [Fe(en)(bpy)(NH 3) 2] 2+, where gly = glycine, en = ethylenediamine and, bpy = bipyridyl moities. (At. nos. Ti = 22, V = 23,, Fe = 26, Co = 27), (2008), 59. In which of the following coordination entities the, magnitude of o (CFSE in octahedral field) will be, maximum?, (a) [Co(CN) 6] 3–, (b) [Co(C 2O4) 3] 3–, 3+, (c) [Co(H2O) 6], (d) [Co(NH 3) 6] 3+, (At. No. Co = 27), (2008), 60. The d electron configurations of Cr2+, Mn2+, Fe2+, and Ni2+ are 3d4, 3d5, 3d6 and 3d8 respectively., Which one of the following aqua complexes will, exhibit the minimum paramagnetic behaviour?, (a) [Fe(H 2O) 6] 2+, (b) [Ni(H 2O) 6] 2+, (c) [Cr(H 2O) 6] 2+, (d) [Mn(H 2O) 6] 2+., (At. No. Cr = 24, Mn = 25, Fe = 26, Ni = 28), (2007), , www.mediit.in

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67, , Coordination Compounds, , 61. [Cr(H2O)6]Cl3 (At. no. of Cr = 24) has a magnetic, moment of 3.83 B.M. The correct distribution of 3d, electrons, of the complex is, 1 in1the chromium, 1, (a) 3d , 3d , 3d, xy, , (b) 3d1 2, , yz, , 2,, , (x  y ), 1, 1, , (c) 3d , 3d, xy, 1, , z2, , 4, 2–, , (d) [NiCl4] - tetrahedral, paramagnetic, , 3d1z2, 3d1 xz, , 9.6, , 1, , 2, , 2 , 3d, , (x  y ), 1, 1, , yz, , (d) 3d , 3d , 3d, xy, , yz, , (2006), , xz, , 62. Which one of the following is an inner orbital, complex as well as diamagnetic in behaviour?, (a) [Zn(NH 3) 6] 2+, (b) [Cr(NH 3) 6] 3+, 3+, (c) [Co(NH 3) 6], (d) [Ni(NH 3) 6] 2+, (Atomic number : Zn = 30, Cr = 24, Co = 27,, Ni = 28), (2005), 63. Among [Ni(CO)4], [Ni(CN)4]2–, [NiCl4]2– species,, the hybridisation states at the Ni atom are,, respectively, (a) sp3, dsp2, dsp2, (b) sp3, dsp2, sp3, (c) sp3, sp3, dsp2, (d) dsp2, sp3, sp3., (2004), , [Atomic number of Ni = 28], –, , 64. CN, a strongnegative, field ligand., This is due to the fact that, (a) itiscarries, charge, (b) it is a pseudohalide, (c) it can accept electrons from metal species, (d) it forms high spin complexes with metal, species., (2004), 65. Considering H2O as a weak field ligand, the number, of unpaired electrons in [Mn(H2O)6]2+ will be, (atomic number of Mn = 25), (a) three, (b) five, (c) two, (d) four., (2004), 66. In an octahedral structure, the pair of d orbitals, involved in d2sp3 hybridisation is, (a) d 2 2 , d 2, (b) dxz , d 2 2, x y, , (c) d 2 , dxz, z, , x y, , z, , (d) dxy , dyz., , (2004), , 67. The number of unpaired electrons in the complex, ion [CoF 6]3– is, (a) 2, (b) 3, (c) 4, (d) zero, (Atomic no. : Co = 27), , (2003), , 68. Atomic number of Cr and Fe are respectively 24 and, 26, which of the following is paramagnetic with the, spin of electron?, (a) [Cr(CO)6], (b) [Fe(CO)5], (c) [Fe(CN) 6] 4–, (d) [Cr(NH 3) 6] 3+ (2002), , www.neetujee.com, , 69. Which statement is incorrect?, (a) Ni(CO) 4 - tetrahedral, paramagnetic, (b) [Ni(CN) 4] 2– - square planar, diamagnetic, (c) Ni(CO) - tetrahedral, diamagnetic, (2001), , Bonding in Metal Carbonyls, , 70. Iron carbonyl, Fe(CO)5 is, (a) tetranuclear, (b) mononuclear, (c) trinuclear, (d) dinuclear., (NEET 2018), 71. An example of a sigma bonded organometallic, compound is, (a) Grignard’s reagent (b) ferrocene, (c) cobaltocene, (d) ruthenocene., (NEET 2017), 72. Which of the following has longest C—O bond, length? (Free C—O bond length in CO is 1.128 Å.), (a) [Fe(CO)4]2–, (b) [Mn(CO)6]+, (c) Ni(CO) 4, (d) [Co(CO) 4]–, (NEET-I 2016), 73. Which of the following carbonyls will have the, strongest C – O, bond?, (a) Mn(CO) +, (b) Cr(CO) 6, (c) V(CO)6–, , 6, , (d) Fe(CO)5, , (2011), , 74. Which of the following does not have a metal carbon bond?, (a) Al(OC2H5)3, (b) C2H5MgBr, (c) K[Pt(C2H4)Cl3], (d) Ni(CO)4, (2004), 75. Among the following which is not the -bonded, organometallic compound?, (a) K [PtCl3(2 – C2H4)], (b) Fe (5 – C5H5)2, (c) Cr(6 – C6 H6) 2, (d) (CH3)4Sn, (2003), 76. Which of the following organometallic compounds, is  and -bonded?, (a) [Fe(5 – C5H5)2], (b) K[PtCl3(2 – C2H4)], (c) [Co(CO) 5NH 3] 2+, (d) Fe(CH3)3, (2001), 77. Shape of Fe(CO)5 is, (a) octahedral, (b) square planar, (c) trigonal bipyramidal (d) square pyramidal., (2000), 78. In metal carbonyl having general formula M(CO)x, where M = metal, x = 4 and the metal is bonded to, (a) carbon and oxygen (b) C O, (c) oxygen, (d) carbon., (1995), , www.mediit.in

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68, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 9.7, , Importance and Applications, Coordination Compounds, , of, , 79. Which of the following complexes is used to be as an, anticancer agent?, (a) mer-[Co(NH ) Cl ] (b) cis-[PtCl (NH ) ], 333, 2, 32, (c) cis, -K2[PtCl2Br2], (d) Na2CoCl4, (2014), 80. Copper sulphate dissolves in excess of KCN to give, (a) Cu(CN) 2, (b) CuCN, (c) [Cu(CN)4]3–, (d) [Cu(CN)4]2–, (2006), 81. Which of the following is considered to be an, anticancer species?, (a), , (b), , (c), , (d), , (2004), , 82. In the silver plating of copper, K[Ag(CN)2] is used, instead of AgNO3. The reason is, (a) a thin layer of Ag is formed on Cu, (b) more voltage is required, (c) Ag+ ions are completely, removed from solution, (d) less availability of Ag+ ions, as Cu cannot, displace Ag from [Ag(CN)2]– ion., , (2002), , 83. CuSO 4 when reacts with KCN forms CuCN, which, is insoluble in water. It is soluble in excess of KCN,, due to formation of the following complex, (a) K2[Cu(CN)4], (b) K3[Cu(CN)4], (c) CuCN 2, (d) Cu[KCu(CN)4], (2002), 84. Hypo is used in photography to, (a) reduce AgBr grains to metallic silver, (b) convert metallic silver to silver salt, (c) remove undecomposed silver bromide as a, soluble complex, (d) remove reduced silver., (1988), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., 61., 71., 81., , (b), (c), (a), (d), (b), (b), (d), (a), (c), , 2., 12., 22., 32., 42., 52., 62., 72., 82., , (c), (d), (b), (c), (d), (c), (c), (a), (d), , 3., 13., 23., 33., 43., 53., 63., 73., 83., , (c), (c), (c), (d), (b), (b), (b), (a), (b), , 4., 14., 24., 34., 44., 54., 64., 74., 84., , (a), (b), (a), (a), (c), (a), (b), (a), (c), , 5., 15., 25., 35., 45., 55., 65., 75., , (a), (a), (b), (c), (b), (d), (b), (d), , 6., 16., 26., 36., 46., 56., 66., 76., , (b), (a), (c), (a), (c), (b), (a), (c), , 7., 17., 27., 37., 47., 57., 67., 77., , (d), (d), (c), (d), (b), (b), (c), (c), , 8., 18., 28., 38., 48., 58., 68., 78., , (c), (a), (d), (b), (a), (a), (d), (d), , 9., 19., 29., 39., 49., 59., 69., 79., , (b), (b), (d), (d), (b), (a), (a), (b), , 10., 20., 30., 40., 50., 60., 70., 80., , (b), (c), (c), (b), (c), (b), (b), (c), , Hints & Explanations, (b) : [Co(NH3)6]Cl3 + 3AgNO3  3AgCl, + [Co(NH3)6](NO3)3, [Co(NH3)5Cl]Cl2 + 2AgNO3  2AgCl, + [Co(NH3)5Cl](NO3)2, [Co(NH3)4Cl2]Cl + AgNO3  AgCl, + [Co(NH 3)4Cl2]NO 3, , 1., , 2. (c) : For octahedral complexes, coordination number, is 6. Hence, CoCl3 · 3NH3 i.e., [Co(NH3)3Cl3] will not, ionise and will not give test for Cl– ion with silver nitrate., , www.neetujee.com, , 3., , (c) : [Cr(H2O)4Cl2]Cl + AgNO3 , [Cr(H2O)4Cl2]NO3 + AgCl, ppt., , No. of millimoles of solution = 100 mL × 0.01 M, = 1 millimole, = 10–3 mole, So, mole of AgCl = 0.001, 4. (a) : Ionic conductance increases with increasing, the number of ions, produced after decomposition., , www.mediit.in

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69, , Coordination Compounds, , Compound, No. of ions produced, K4[Fe(CN)6], 5, [Co(NH3)6]Cl3, 4, [Cu(NH3)4]Cl2, 3, [Ni(CO)4], 0, 5. (a) : As the complex gives two moles of AgCl ppt., with AgNO3 solution, so the complex must have two, ionisable Cl atoms. Hence, the probable complex, which, gives three mole ions may be [Co(NH3)5NO2]Cl2., [Co(NH3)5NO2]Cl2  [Co(NH3)5NO2]2+ + 2Cl–, one mole  3 mole ions, 6. (b) : The intensity of the trans-effect (as measured, by the increase in rate of substitution of the trans ligand), follows the sequence : CN– > C H– > Br– > NH, 6, , 5, , 3, , 7. (d) : [M(en)2(C2O4)]Cl :, Oxidation number of metal = +3, Coordination number of metal = 6,  Sum of oxidation number and coordination number, = 3 + 6 =9, 8. (c) : The ligand acetylacetone forms six membered, chelate ring in the complex [Co(acac)3]., 9. (b) : Bond energy of F2 is less than Cl2 due to interelectronic repulsions in small sized F-atoms., Silicon exhibits coordination number 6., In aqueous state, Mn(II) is more stable., , Mn Mn2  2e, The common oxidation states of 15th group elements are, –3, +3 and +5., 10. (b) : C O2–  bidentate ligand., 2, , 13. (c), 14. (b), 15. (a) : Chlorodiaquatriamminecobalt(III) chloride, can be represented as [CoCl(NH3)3(H2O)2]Cl2., 16. (a) : The ligands are named in the alphabetic order, according to latest IUPAC system. So, the name of, [Pt(NH 3) 3Br(NO 2)Cl]Cl is triamminebromochloronitro, platinum(IV) chloride. (The oxidation no. of ‘Pt’ is +4)., 17. (d) : The formula of dichlorobis(urea)copper(II) is, [CuCl2{(NH2)2CO}2]., 18. (a) : [CoCl (en) 2], exhibits geometrical isomerism,, 2, , as the coordination number of Co is 6 and this compound, has octahedral geometry., , 19. (b) : Possible isomers of [Co(en)2Cl2]Cl :, , en, , en, en, , en, cis, l, d, , 4, , 3 molecules attached from two sides with Ni makes, coordination number 6., 11. (c) :, en, , +3, , As the number of atoms of the ligands that are directly, bound to the central metal is known as coordination, number. It is six here (see in figure)., Oxidation state : Let oxidation state of Cr be x.,  3 (+1) + x + 3 (–2) = 0  3 + x – 6 = 0  x = + 3, 12. (d) : When a ligand has two groups that are capable, of bonding to the central atom, it is said to be bidentate., Thus, the only ligand,which is expected to be bidentate is, C2O4 2– as, , www.neetujee.com, , en, , trans, , 20. (c) : Coordination isomerism arises from the, interchange of ligands between cationic and anionic, entities of different metal ions present in the complex., e.g., [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6], 21. (a) : [Pt(py)(NH3)BrCl] can have three isomers., , www.mediit.in

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70, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 22. (b) :, +, , Cl, , H3N, , Cl, Co, , H3 N, , +, , Cl, H3N, , NH3, Co, , NH3, NH3, cis-form, (violet in colour), , H3 N, , NH3, Cl, trans-form, (green in colour), , cis-form, , 23. (c) : Compounds having tetrahedral geometry does, not exhibit isomerism due to presence of symmetry, elements. Here, [Ni(NH3)2Cl2] has tetrahedral geometry., 24. (a) : Optical isomerism is shown by :, (i) Complexes of the type [M(AA) 2Y2], containing one, symmetrical bidentate ligand i.e., [Co(en)Cl2(NH3)2]+., (ii) Complexes of the type [M(AA)3], containing a symmetrical bidentate ligand i.e., [Co(en)3]3+., (iii) Complexesofthetype [M(AA)2X2], i.e., [Co(en)2Cl2]+., However complexes of the type [MA3B3] show geometrical, isomerism, known as fac-mer isomerism.,  [Co(NH3)3Cl3] exhibits fac-mer isomerism., , trans-form, , 27. (c) : Optical isomerism is not shown by square, planar complexes., Octahedral complexes of general formulae,, [Ma2b2c2]n±, [Mabcdef], [M(AA)3]n±, [M(AA)2a2]n± (where, AA = symmetrical bidentate ligand), [M(AA)2ab]n± and, [M(AB)3]n±, (where AB = unsymmetrical ligands) show optical, isomerism., , trans-[Co(en)2Cl2]+, , does not show optical isomerism (superimposable mirror, image). But cis-form shows optical isomerism., , fac, , mer, , 25. (b) : Either a pair of crystals, molecules or, compounds that are mirror images of each other but, are not identical, and that rotate the plane of polarised, light equally, but in opposite directions are called as, enantiomorphs., , 28. (d) : [Co(en)3]3+ :, , 26. (c) : Ionization isomerism arises when the, coordination compounds give different ions in solution., –, [Co(NH3)4(NO2)2]Cl  [Co(NH3)4(NO2)2]+ + Cl, [Co(NH3)4(NO2)Cl]NO2  [Co(NH3)4(NO2)Cl]+ + NO–, , 29. (d) : [Cr(SCN)2(NH3)4]+ shows linkage, geometrical, and optical isomerism., 30. (c) : [Fe(PPh ) NH ClBr]Cl can give two optical, , 2, , Linkage isomerism occurs in complex compounds which, contain ambidentate ligands like NO2–, SCN–, CN–,, S2O32– and CO., [Co(NH 3) 4(NO2) 2]Cl and [Co(NH 3) 4(ONO) 2]Cl, are linkage isomers as NO2– is linked through N or, through O., Octahedral complexes of the type Ma4b2 exhibit, geometrical isomerism., , www.neetujee.com, , 3 3, , 3, , and two geometrical isomers. While other complexes do, not form geometrical isomers., 31. (d) : The isomers of the complex compound, [CuII(NH3)4][PtIICl4] are :, (i) [Cu(NH 3) 3Cl] [Pt(NH 3)Cl3], (ii) [Pt(NH 3) 3Cl][Cu(NH 3)Cl3], (iii) [Pt(NH 3) 4][CuCl4], So, the total no. of isomers are = 4, , www.mediit.in

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71, , Coordination Compounds, , 32. (c) : Possible geometrical isomers are :, NH3, NH3, O2N, , NH3, , H 3N, , Co, , NO2, Co, , NH3, , O 2N, , NO2, , O 2N, NH3, , NO2, fac-, , mer–, , 33. (d) :, , cis-form, trans-form, 34. (a) : According to spectrochemical series, order of, increasing– field– strength, is :, SCN < F < C O 2– < CN–, 24, , 35. (c) : In K4[Fe(CN)6] complex, Fe is in +2 oxidation, state., 3d, , 40. (b) : [Mn(CN)6]3– : Let oxidation state of Mn be x., x + 6 × (–1) = – 3  x = +3, Electronic configuration of Mn : [Ar]4s2 3d5, Electronic configuration of Mn3+ : [Ar]3d4, CN– is a strong field ligand thus, it causes pairing of, electrons in 3d-orbital., , 4s, , Fe2+ :, As CN– is a strong field ligand, it causes pairing of, electrons therefore, electronic configuration of Fe2+ in, K4[Fe(CN)6] is t2g6ge0., 36. (a), 4, 4, =  18000  8000 cm1, 37. (d) : t  , o, 9, 9, 38. (b) : Ni(28) : [Ar]3d84s2, , Thus, [Mn(CN)6]3– has d2sp3 hybridisation and has, octahedral geometry., 41. (b) : Jahn–Teller distortion is usually significant, for asymmetrically occupied eg orbitals since they, are directed towards the ligands and the energy gain is, considerably more., In case of unevenly occupied t2g orbitals, the Jahn–Teller, distortion is very weak since the t2g set does not point, directly at the ligands and therefore, the energy gain is, much less., High spin complexes :, , d4, unsymmetrical d7, unsymmetrical d8, symmetrical d9, unsymmetrical, , 42. (d) : [Ni(CN)4]2– : Oxidation number of Ni = +2, Electronic configuration of Ni2+ : 3d84s0, 3d, , [Ni(CN)4]2, , 4s, , 4p, ,  CO is a strong field ligand, so, unpaired electrons, get paired., In [Ni(CO)4] :, , Thus, the complex is sp3 hybridised with tetrahedral, geometry and diamagnetic in nature., 39. (d) : Increasing order of crystal field splitting energy, is : H2O < NH3 < en, Thus, increasing order of crystal field splitting energy for, the given complexes is :, [Co(H2O)6]3+ < [Co(NH3)6]3+ < [Co(en)3]3+, hc, As, E , , Thus, increasing order of wavelength of absorption is :, [Co(en) 3] 3+ < [Co(NH 3) 6] 3+ < [Co(H2O) 6] 3+, , www.neetujee.com, , Pairing of electrons in d-orbital takes place due to the, presence of strong field ligand (CN–)., 43. (b) : H2O is a weak field ligand, hence o < pairing, energy., CFSE = (–0.4x + 0.6y)o, where, x and y are no. of electrons occupying t2g and eg, orbitals respectively., 3+, For3+[Fe(H, 2O)6] complex ion,, Fe (3d5) = t3 e 2 = – 0.4 × 3 + 0.6 × 2 = 0.0 or 0 , 2g g, , 0, , 44. (c) : Oxidation state of Cu in [Cu(NH3)4]2+ is + 2, Cu2+ = 3d9, It has one unpaired electron (n = 1)., , , , n(n  2)  1(1 2)  3  1.73 BM, , 45. (b) : CFSE = (– 0.4 x + 0.6 y) o, where, x = No. of electrons occupying t2g orbitals, y = no. of electrons occupying eg orbitals, = (– 0.4 × 3 + 0.6 × 1)o [ High spin d4 = t3 2ge1] g, = (– 1.2 + 0.6)o = –0.6 o, , www.mediit.in

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73, , Coordination Compounds, , 58. (a) : O.S. of Ti in the complex [Ti(NH3)6]3+ is +3.,  14.6689 = n2 + 2n, 3+, 2, 2, 6, 2, 6, 1, Ti : 1s 2s 2p 3s 3p 3d, On solving the equation, n = 3, Cr3+  [Ar]3d3, No. of unpaired electrons in d orbital is one., +, Let O.S. of V in complex [V(gly)2(OH)2(NH3)2] is x., x + 2 × 0 + 2 × (–1) + 2 × 0 = +1,  x = +3, 62. (c) : [Co(NH3)6]3+; Co(27) : [Ar]18 3d7 4s2, V3+ : 1s2 2s2 2p6 3s2 3p6 3d2, No. of unpaired electrons in d orbital is two., O.S. of Fe in complex [Fe(en)(bpy)(NH3)2]2+ is +2.,  Fe2+ : 1s2 2s2 2p6 3s2 3p6 3d6, Co3+ : [Ar]18 3d6, As all are strong ligands so, pairing of electrons takes, place., No. of unpaired electron in d orbital is zero., Let O.S. of Co in the given complex [Co(ox)2(OH)2]– is x., d2sp3, x + 2 × (–2) + 2 × (–1) = –1  x – 4 – 2 = –1, electron pair from six ligands(NH3),  x = +5, d2sp3 inner octahedral complex and diamagnetic., Co5+ : 1s2 2s2 2p6 3s2 3p6 3d4, [Zn(NH3)6]2+  sp3d2(outer)and diamagnetic., –, As ox and OH are weak field ligands so no pairing of, [Cr(NH3)6]3+  d2sp3(inner)and paramagnetic., 3 1, electrons takes place, t2g e gso, it has 4 unpaired electrons [Ni(NH3)6]2+  sp3d2(outer)and paramagnetic., and has highest paramagnetic behaviour., 63. (b) :, 59. (a) : When the ligands are arranged in order of the, magnitude of crystal field splitting, the arrangement,, :, thus, obtained is called spectrochemical series., Arranged, in increasing field strength as, :, I– < Br– < Cl– < NO3– < F– < OH– < C O 2– < H O < NH, –, –, 2 4, 2, 3, < en, < NO2 < CN < CO, It has been observed that ligands before H2O are weak field, ligands while ligands after H2O are strong field ligands., :, , , as CN is a strong eld ligand., , :, CFSE in octahedral field depends upon the nature of, ligands. Stronger the ligands larger will be the value of, oct., 60. (b) : 3d, 4 unpaired electrons, 5 unpaired electrons, 4 unpaired electrons, 2 unpaired electrons, Greater the number of unpaired electrons, higher is the, paramagnetism. Hence, [Ni(H2O)6]2+ will exhibit the, minimum paramagnetic behaviour., 61. (d) : Magnetic moment , n(n  2), , 3.83  n(n  2)  (3.83)2  n(n  2), www.neetujee.com, , as Cl is a weak eld ligand., , 64. (b) : Cyanide ion is strong field ligand because it, is a pseudohalide ion. Pseudohalide ions are stronger, coordinating ligands and they have the ability to form, -bond (from the pseudohalide to the metal) and  bond, (from the metal to pseudohalide)., 65. (b) : Mn (25)  3d54s2, :, In presence of weak field ligand, there will be, no pairing of electrons. So, it will form a high spin, complex, i.e., the number of unpaired electrons = 5., 66. (a) : In the formation of d2sp3 hybrid orbitals, two, (n – 1)d orbitals of eg set [i.e. (n – 1)dz2 and (n – 1)d x2 2–y, orbitals)], one ns and three np(npx, npy and npz) orbitals, combine together and form six d2sp3 hybrid orbitals., , www.mediit.in

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74, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, 3–, , 67. (c) : [CoF6] :, , :, :, Thus, the number of unpaired electrons = 4., 68. (d) : Odd electrons, ions and molecules are, paramagnetic., In Cr(CO) 6 molecule 12 electrons are contributed by CO, group and it contains no odd electron., Cr : 3d5 4s1, Fe(CO)5 molecule also does not contain odd electron., Fe : 3d6 4s2, In [Fe(CN)6]4– ion Fe(+2) : 3d6 4s0,  No odd electrons., In [Cr(NH3)6]3+ ion Cr(+3) : 3d3 4s0, This ion contains odd electron so it is paramagnetic., 69. (a) : In Ni(CO)4 complex, Ni(0) will have 3d10, configuration., , Hence, [Ni(CO)4] will have tetrahedral geometry and, diamagnetic as there are no unpaired electrons., 70. (b) : Based on the number of metal atoms present in, a complex, they are classified as :, e.g., : Fe(CO)5 : mononuclear, Co2(CO)8 : dinuclear ; Fe3(CO)12 : trinuclear, 71. (a) : In sigma bonded organometallic complexes,, the metal atom and carbon atom of the ligand are joined, together with a sigma bond, i.e., ligand contributes one, electron and is therefore, called one electrons donor, e.g.,, Grignard’s reagent R-Mg-X., 72. (a) : The greater the negative charge on the carbonyl, complex, the more easy it would be for the metal to permit, its electrons to participate in the back bonding, the higher, would be the M—C bond order and simultaneously there, would be larger reduction in the C—O bond order. Thus,, [Fe(CO)4]2– has the lowest C—O bond order means the, longest bond length., 73. (a) : The presence of positive charge on the metal, carbonyl would resist the flow of the metal electron, charge to * orbitals of CO. This would increase the CO, bond order and hence, CO in a metal carbonyl cation, would absorb at a higher frequency compared to its, absorption in a neutral metal carbonyl., 74. (a) : Al(OC 2H5) 3 contains bonding through O and, thus it does not have metal - carbon bond., , 75. (d) : -bonded organometallic compound includes, organometallic compounds of alkenes, alkynes and some, other carbon containing compounds having electrons in, their p-orbitals., 76. (c) : [Co(CO) 5NH 3] 2+ : In this complex, Co-atom, is attached with NH 3 through  bonding and with CO, through dative -bond., 77. (c) : In Fe(CO)5, the ‘Fe’ atom is dsp3 hybridised,, therefore, shape of the molecule is trigonal bipyramidal., Fe atom in Fe(CO)5, , 78. (d) : In M(CO)4, metal is bonded to the ligands via, carbon atoms with both  and -bond character. Both, metal to ligand and ligand to metal bonding are possible., 79. (b), 80. (c) : First cupric cyanide is formed which, decomposes to give cuprous cyanide and cyanogen gas., Cuprous cyanide dissolves in excess of potassium cyanide, to form a complex, potassium cyanide [K3Cu(CN)4]., [CuSO4 + 2KCN  Cu(CN)2 + K2SO4] × 2, 2Cu(CN)2  Cu2(CN)2 + (CN)2, Cu2(CN)2 + 6KCN  2K3Cu(CN)4, 2CuSO4 + 10KCN  2K3Cu(CN)4 + 2K2SO4 + (CN)2, 81. (c) : cis-platin, is cis-[PtCl (NH ) ] is used as an, 2, , 32, , anticancer agent., 82. (d) : Copper being more electropositive readily, precipitate silver from their salt (Ag+) solution., Cu + 2AgNO3  Cu(NO3)2 + Ag, –, In K[Ag(CN)2] solution, a complex anion [Ag(CN)2] is, formed so Ag+ ions are less available in the solution and, Cu cannot displace Ag from this complex ion., 83. (b) : Copper sulphate reacts with potassium cyanide, giving a white precipitate of cuprous cyanide and, cyanogen gas. The cuprous cyanide dissolves in excess of, KCN forming potassium cuprocyanide K3[Cu(CN)4]., 2CuSO4 + 4KCN  2CuCN + (CN)2 + 2K2SO4, CuCN + 3KCN  K3[Cu(CN)4], 84. (c) : Undecomposed AgBr forms a soluble complex, with hypo and the reaction is given as :, AgBr + 2Na2S2O3  Na3[Ag(S2O3)2] + NaBr, Soluble complex, , , www.neetujee.com, , www.mediit.in

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76, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (C), (b), , (c), , (NEET 2017), , (d), 9., , Consider the reaction,, CH3CH2CH2Br + NaCN, , CH3CH2CH2CN + NaBr, This reaction will be the fastest in, (a) ethanol, (b) methanol, (c) N , N  -dimethylformamide (DMF), (d) water., (NEET-II 2016), 10. Which of the following biphenyls is optically active?, , (b), , (c), , 13. In an SN1 reaction on chiral centres, there is, (a) inversion more than retention leading to partial, racemisation, (b) 100% retention, (c) 100% inversion, (d) 100% racemisation., (2015), , (i), , (ii) CH3CH2CH2Cl, , (iii), , (iv), , (a) (i) and (ii), (c) (iii) and (iv), , (b) (ii) and (iv), (d) (i) and (iv), , 15. Given:, , (2014), , (NEET-I 2016), , (d), , 11. For the following reactions :, (A) CH CH CH Br + KOH, 3, , 2, , 2, , CH3CH, , www.neetujee.com, , 12. Twopossiblestereo-structuresof CH3CHOHCOOH,, which are optically active, are called, (a) atropisomers, (b) enantiomers, (c) mesomers, (d) diastereomers. (2015), , 14. Which of the following compounds will undergo, racemisation when solution of KOH hydrolyses?, , (a), , (B), , Which of the following statements is correct?, (a) (A) is elimination, (B) and (C) are substitution, reactions., (b) (A) is substitution, (B) and (C) are addition, reactions., (c) (A) and (B) are elimination reactions and (C) is, addition reaction., (d) (A) is elimination, (B) is substitution and (C) is, addition reaction., (NEET-I 2016), , CH2 + KBr + H2O, , I and II are, (a) identical, (b) a pair of conformers, (c) a pair of geometrical isomers, (d) a pair of optical isomers., (Karnataka NEET 2013), 16. Which of the following acids does not exhibit optical, isomerism?, (a) Maleic acid, (b) -Amino acids, (c) Lactic acid, (d) Tartaric acid (2012), , www.mediit.in

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78, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 34., , 29. The chirality of the compound, , n-butane will be, (a) meso form, (c) d-form, , is, (a) R, (c) E, , obtained by chlorination of, , (b) S, (d) Z, , (2005), , 30. Which of the following is least reactive in a, nucleophilic substitution reaction?, (a) (CH3)3C – Cl, (b) CH2 CHCl, (c) CH3CH2Cl, (d) CH2 CHCH2Cl, (2004), 31. Which of the following pairs of compounds are, enantiomers?, , (a), , (b) racemic mixture, (d) l-form., (2001), , 35. An organic compound A(C4H9Cl) on reaction with, Na/diethyl ether gives a hydrocarbon which on, monochlorination gives only one chloro derivative, then, A is, (a) t-butyl chloride (b) s-butyl chloride, (c) iso-butyl chloride (d) n-butyl chloride., (2001), 36. A compound of molecular formula C7H16 shows, optical isomerism, compound will be, (a) 2,3-dimethylpentane, (b) 2,2-dimethylbutane, (c) 2-methylhexane, (d) none of these., (2001), 37. Which of the following compounds is not chiral?, (a) CH3CHDCH2Cl (b) CH3CH2CHDCl, (c) DCH2CH2CH2Cl (d) CH3CHClCH2D, (1998), , (b), , 38. Replacement of Cl of chlorobenzene to give, phenol requires drastic conditions. But chlorine, of 2,4-dinitrochlorobenzene is readily replaced, because, (a) NO2 donates e– at meta position, (b) NO2 withdraws e– from ortho/para positions, (c) NO2 makes ring electron rich at ortho and para, (d) NO2 withdraws e– from meta position., (1997), , (c), , (d), (2003), 32. Reactivity order of halides for dehydrohalogenation, is, (a) R – F > R – Cl > R – Br > R – I, (b) R – I > R – Br > R – Cl > R – F, (c) R – I > R – Cl > R – Br > R – F, (d) R – F > R – I > R – Br > R – Cl, (2002), NaC N, , Ni/H, , 2, 33. CH3CH2Cl   X  , , acetic anhydride, , Y      Z, Z in the above reaction sequence is, (a) CH3CH2CH2NHCOCH3, (b) CH3CH2CH2NH2, (c) CH3CH2CH2CONHCH3, (d) CH3CH2CH2CONHCOCH3, (2002), , www.neetujee.com, , 39. The alkyl halide is converted into an alcohol by, (a) elimination, (b) dehydrohalogenation, (c) addition, (d) substitution., , (1997), , 40. The following reaction is described as, , (a) SN2, (c) SE2, , (b) SN0, (d) SN1, , (1997), , 41. Reaction of t-butyl bromide with sodium methoxide, produces, (a) sodium t-butoxide, (b) t-butyl methyl ether, (c) isobutane, (d) isobutylene., (1994), , www.mediit.in

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Haloalkanes and Haloarenes, , 79, , 42. Grignard reagent is prepared by the reaction, between, (a) magnesium and alkane, (b) magnesium and aromatic hydrocarbon, (c) zinc and alkyl halide, (d) magnesium and alkyl halide., , 10.8 Polyhalogen Compounds, , (1994), , 47. Trichloroacetaldehyde, CCl3CHO reacts with, chlorobenzene in presence of sulphuric acid and, produces, Cl, , 43. Chlorobenzene reacts with Mg in dry ether to give a, compound (A) which further reacts with ethanol to, yield, (a) phenol, (b) benzene, (c) ethyl benzene, (d) phenyl ether. (1993), , (a) Cl, , OH, , 44. Benzene reacts with n-propyl chloride in the, presence of anhydrous AlCl3 to give, (a) 3-propyl-1-chlorobenzene, (b) n-propylbenzene, (c) no reaction, (d) isopropylbenzene., , (b) Cl, , Cl, CH, , 45. Which chloro derivative of benzene among the, following would undergo hydrolysis most readily, with aqueous sodium hydroxide to furnish the, corresponding hydroxy derivative?, , (c), , (d) C6H5Cl, , (1989), , 46. Which of the following is an optically active, compound?, (a) 1-Butanol, (b) 1-Propanol, (c) 2-Chlorobutane, (d) 4-Hydroxyheptane, (1989), , Cl, , CCl3, , (1993), , (b), , Cl, , C, , (c) Cl, , (a), , Cl, , C, H, , Cl, C, , (d) Cl, , Cl, (2009), , CH2Cl, , 48. Industrial preparation of chloroform employs, acetone and, (a) phosgene, (b) calcium hypochlorite, (c) chlorine gas, (d) sodium chloride., (1993), 49. Phosgene is a common name for, (a) phosphoryl chloride, (b) thionyl chloride, (c) carbon dioxide and phosphine, (d) carbonyl chloride., , (1988), , ANSWER KEY, , 1., 11., 21., 31., 41., , (c), (d), (c), (a), (d), , www.neetujee.com, , 2., 12., 22., 32., 42., , (c), (b), (d), (b), (d), , 3., 13., 23., 33., 43., , (b), (a), (c), (a), (b), , 4., 14., 24., 34., 44., , (a), 5., (None) 15., (b), 25., (b), 35., (d), 45., , (a), (b), (a), (a), (a), , 6., 16., 26., 36., 46., , (d), (a), (b), (a), (c), , 7., 17., 27., 37., 47., , (a), (c), (d), (c), (c), , 8., 18., 28., 38., 48., , (d), (a), (d), (b), (b), , 9., 19., 29., 39., 49., , (c), (c), (a), (d), (d), , 10., 20., 30., 40., , (d), (a), (b), (a), , www.mediit.in

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80, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , Hints & Explanations, 1., , (c) :, , HBr, , CH CH CH3, Br, , 7., , +, , –, , CH, , (a) :, , CH2 CH3, , More stable, (Benzyl carbocation), , CH CH2, , CH3, , Br, 2. (c) : Tertiary halide shows SN1 mechanism i.e., ionic, mechanism. In the given reaction negative ion will attack, on carbocation. Thus greater the tendency of ionisation, (greater ionic character in M – F bond) more favourable, will be reaction. The most ionic bond is Rb – F in the, given examples thus most favourable reaction will be, with Rb–F., 3., , (b) :, , At 400°C temperature, substitution occurs instead of, addition., Br, 5, , 4., , 8. (d) : m-Bromoanisole gives only the respective meta, substituted aniline. This is a substitution reaction which, goes by an elimination-addition pathway., , (a) : CH3, , 4, , 3, , 2, , 1, , CH2, , CH2, , CH, , CH3, , 2-Bromopentane, , -Elimination, , (–HBr), dehydrobromination, , 9. (c) : The reaction,, CH3CH2CH2Br + NaCN, , CH3CH2CH2CN + NaBr, , follows SN2 mechanism which is favoured by polar, aprotic solvent i.e., N , N  dimethylformamide (DMF),, , 4, , 3, , 2, , 1, , CH2, , CH, , CH, , CH3, , 10. (d) : o-Substituted biphenyls are optically active, as both the rings are not in one plane and their mirror, images are non-superimposable., , 5. (a) : Arylhalidesarelessreactiveascomparedtoalkyl, halides as the halogen atom in these compounds is firmly, attached and cannot be replaced by nucleophiles such as, OH–, NH–2, etc. In chlorobenzene, the electron pair, of chlorine atom is in conjugation with -electrons of, benzene ring. Thus C—Cl bond acquires double bond, character and is difficult to break., , 11. (d) : CH3CH2CH2Br + KOH CH3CH CH2, + KBr + H2O, Saturated compound is converted into unsaturated, compound by removal of group of atoms hence, it is an, elimination reaction., , 6., , —Br group is replaced by — OH group hence, it is a, substitution reaction., , 5, , CH3, , Pent-2-ene, (More substituted alkene, is formed, Zaitsev’s rule), , (d) :, , C H O Na   C H Cl N   C H OC H, S 2, , 2 5, (B), , www.neetujee.com, , 2 5, (C), , Williamson’s, synthesis, , 2 5, 2 5, Diethyl ether, , Addition of Br 2 converts an unsaturated compound, into a saturated compound hence, it is an addition, reaction., , www.mediit.in

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Haloalkanes and Haloarenes, , 81, , 12. (b) :, , 13. (a) : In case of optically active alkyl halides,, SN1 reaction is accompanied by racemisation. The, carbocation formed in the slow step being sp2 hybridised, is planar and attack of nucleophile may take place from, either side resulting in a mixture of products, one having, the same configuration and other having inverted, configuration., The isomer corresponding to inversion is present in, slight excess because SN1 also depends upon the degree, of shielding of the front side of the reacting carbon., , C—X bond. The 3° carbocation (formed from III) will, be more stable than its 2° counter part (formed from IV), which in turn will be more stable than the arenium ion, (formed from I). Also, the aryl halide has a double bond, character in the C—X bond which makes the cleavage, more difficult. However, inspite of all the stated factors,, II will be more reactive than I due to the presence of the, electron withdrawing –NO2 group. C—X bond becomes, weak and undergoes nucleophilic substitution reaction., 21. (c) :, , 22. (d), 23. (c) : The given compound may be written as, , 14. (None) : Due to chirality, , , only, , compound (iv) will undergo racemisation., Hence, all the given options are incorrect., 15. (b) : I and II are staggered and eclipsed conformers., 16. (a) : Maleic acid shows geometrical isomerism and, not optical isomerism., , 17. (c) : If reaction is SN1, there will be the formation of, carbocation and the rearrangement takes place. In these, reactions there is no rearrangement hence both are SN2, mechanism., 18. (a) : Electron withdrawing groups like — NO2, facilitates nucleophilic substitution reaction in, chlorobenzene., 19. (c) : SN1 reactions proceed via the formation of a, carbocation intermediate., More stable is the carbocation more reactive is the alkyl/, aryl halide towards SN1., In C6H5C+(CH3)(C6H5) carbocation, the two phenyl rings, by their +R effect and —CH3 by its +I effect diminish the, positive charge and make it stable., , Both geometrical isomerism (cis-trans form) and optical, isomerism is possible in the given compound., No. of optical isomer = 2n = 21 = 2, (where n = no. of asymmetric carbon), Hence total no. of stereoisomers = 2 + 2 = 4, 24. (b) : SN2 mechanism is followed in case of primary, and secondary alkyl halides i.e. SN2 reaction is favoured, by small groups on the carbon atoms attached to halogen, so, CH3 – X > R – CH2 – X > R2CH – X > R3C – X., Primary is more reactive than secondary and tertiary, alkyl halides., 25. (a) : Meso compound does not rotate plane, polarised light. Compound which contains tetrahedral, atoms with four different groups but the whole molecule, is achiral, is known as meso compound. It possesses a, plane of symmetry and is optically inactive. One of the, asymmetric carbon atoms turns the plane of polarised, light to the right and other to the left and to the same extent, so that the rotation due to upper half is compensated by, the lower half, i.e., internally compensated, and finally, there is no rotation of plane polarised light., b, , C2H5, 26. (b) : Cl – C – CH3, 20. (a) : I < II < IV < III, The order of reactivity is dependent on the stability of, the intermediate carbocation formed by cleavage of, , www.neetujee.com, , a, , c, H, d, R-configuration, , www.mediit.in

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82, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 27. (d) :, Due to absence of asymmetric carbon atom., 28. (d) : SN1 reaction is favoured by heavy (bulky), groups on the carbon atom attached to halogens and, nature of carbonium ion in substrate is, Benzyl > Allyl > Tertiary > Secondary > Primary >, Methyl halides., , C – I > C – Br > C – Cl > C – F. The order of bond, dissociation energy R – F > R – Cl > R – Br > R – I. During, dehydrohalogenation C – I bond breaks more easily than, C – F bond. So reactivity order of halides is,, R – I > R – Br > R – Cl > R – F., aC N, 33. (a) : CH3CH2 Cl N,  CH3CH2CN, (X ), , Ni/H, , 2,  CH CH CH NH acetic,  ,  , 3, 2, 2, 2, , anhydride, , (Y ), , (1), , CH3CH2CH2NHCOCH3, (Z ), , 29. (a) :, , 34. (b) : Chlorination of n-butane takes place via free, radical formation. i.e., , (3), , , , (2), , Lowest priority atom is always away from the viewer., Priority is seen on the basis of atomic no. and if atomic, no. are same then on the basis of atomic mass., If clockwise then it is R, if anticlockwise then it is S., Name of the molecule is, (R) 1-bromo-1-chloroethane., 30. (b) : The non-reactivity of the chlorine atom in vinyl, chloride can be explained from the molecular orbital, point of view as follows. If the chlorine atom has sp2, hybridisation, the C – Cl bond will be a -bond and the, two lone pairs of electrons would occupy the other two, sp2 orbitals. This would leave a p orbital containing a lone, pair, and this orbital could now conjugate with the -bond, of the ethylenic link. Thus two M.O’s will be required to, accommodate these four -electrons. Furthermore, since, chlorine is more electronegative than carbon, the electrons, will tend to be found in the vicinity of the chlorine atom., Nevertheless, the chlorine atom has now lost full control, of the lone pair and so, is less negative than it would have, been had there been no conjugation. Since two carbon, atoms have acquired a share in the lone pair, each carbon, atom acquires a small negative charge. Hence, owing to, delocalisation of bonds (through conjugation), the vinyl, chloride molecule has an increased stability. Before the, chlorine atom can be displaced by some other group, the, lone pair must be localised again on the chlorine atom. This, requires energy, and so the chlorine is more firmly bound., , 31. (a) : These two are non-superimposable mirror, images of each other, so they are enantiomers., 32. (b) : I > Br > Cl > F atomic radii, F, Cl, Br, I belong to the same group orderly. Atomic, radii go on increasing as the nuclear charge increases in, preceding downwards in a group. The decreasing order, of bond length, , www.neetujee.com, , Cl, CH3CH2CH2CH 3 , , , sp2- hybrid planar shape intermediate and, attack from either side to give, , may, , 35. (a) : Wurtz reaction : It involves the reaction of, alkyl halides with Na in ether to form higher alkanes., 2R – X + 2Na, R – R + 2NaX, In the given problem,, Ether, 2C H Cl  2Na  ,  C H  C H  2NaCl, 49, (A), , 4 9, , 49, , Compound A is t-butyl chloride, in this compound all, — CH3 groups have primary hydrogen only and able to, give only, one chloro derivative., Cl, , (CH3)3 CC(CH3)3 2 CH2 Cl(CH3)2 C– C(CH3)3, 36. (a) : Organic compounds exhibit the property of, enantiomerism (optical isomerism) only when their, molecules are chiral. Most chiral compounds have a, chiral centre, which is an atom bonded to four different, atoms or groups., , 2,3-Dimethylpentane has one chiral C-atom and does, not have any symmetric element., 37. (c) :, , Cl, , www.mediit.in

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Alcohols, Phenols and Ethers, , 85, , (a) three, (b), , (b) four, , (c) five, (d) two., (Karnataka NEET 2013), , 13. In the following sequence of reactions,, H O, , KCN, , , , LiAlH, ether, , 3, CH3  Br , B  4, C,  A , , the end product (C) is, (a) acetone, (b) methane, (c) acetaldehyde, (d) ethyl alcohol., (c), , 14. In the following reactions,, CH3, (i) CH3, , OH, (d) P =, , 8., , (2012), , CH(CH 3) 2, , CH, , CH, , +, CH3 H /heat, , A +, , OH, , Q=, , Major, product, , (NEET 2018), R = CH3 CO CH3, Which one is the most acidic compound?, , , , (ii), , HBr, dark, , A     , in absence of peroxide, , C, , B, Minor, product, , , , D, ,  Major   Minor ,  product  product , , the major products (A) and (C) are respectively, (a), , (b), (a), , (c), , (d), , (NEET 2017), (b), , 9., , Reaction of phenol with chloroform in presence of, dilute sodium hydroxide finally introduces which, one of the following functional group?, (a) –COOH, (b) –CHCl2, (c) –CHO, (d) –CH2Cl, (2015), , 10. Which of the following reaction(s) can be used for, the preparation of alkyl halides?, nh .Zn Cl2, (I) CH 3 CH2 OH + HCl A, , , (II) CH3CH2OH + HCl, (III) (CH3)3COH + HCl, Anh .Zn Cl, (IV) (CH3)2 CHOH + HCl  2, , (a) (I) and (II) only, (b) (IV) only, (c) (III) and (IV) only, (d) (I), (III) and (IV) only, (2015), 11. Which of the following will not be soluble in sodium, hydrogen carbonate?, (a) 2,4,6-Trinitrophenol, (b) Benzoic acid, (c) o-Nitrophenol, (d) Benzenesulphonic acid, (2014), 12. Number of isomeric alcohols of molecular formula, C6H14O which give positive iodoform test is, , www.neetujee.com, , (c), , (d), (2011), 15. Given are cyclohexanol (I), acetic acid (II), 2,4,6trinitrophenol (III) and phenol (IV). In these the, order of decreasing acidic character will be, (a) III > II > IV > I, (b) II > III > I > IV, (c) II > III > IV > I, (d) III > IV > II > I, ( 2010), , www.mediit.in

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86, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 16. Which of the following compounds has the most, acidic nature?, , (a) Cl – CH2 – CH2 – OH, , (a), , (b), , (b), , (c), , (d), , (d), (2010), , 17. Among the following four compounds, (i) Phenol, (ii) Methyl phenol, (iii) Meta-nitrophenol (iv) Para-nitrophenol, The acidity order is, (a) (iv) > (iii) > (i) > (ii) (b) (iii) > (iv) > (i) > (ii), (c) (i) > (iv) > (iii) > (ii) (d) (ii) > (i) > (iii) > (iv), (2010), 18. When glycerol is treated with excess of HI, it, produces, (a) 2-iodopropane, (b) allyl iodide, (c) propene, (d) glycerol triiodide., (Mains 2010), 19. Consider the following reaction :, PBr, , alc.KOH, , Ethanol 3,  X   , Y, (i) H SO , room temperatu re, (ii) H2O, heat, , 2 4,        Z,  , , the product Z is, (a) CH3CH2 – O – CH2 – CH3, (b) CH3 – CH2 – O – SO3H, (c) CH3CH2OH, (d) CH2 CH2, , (2009), , 20. HOCH2CH2OH on heating with periodic acid gives, (a) 2HCOOH, (b), (d) 2CO2, , 25. n-Propyl alcohol and isopropyl alcohol can be, chemically distinguished by which reagent?, (a) PCl5, (b) Reduction, (c) Oxidation with potassium dichromate, (d) Ozonolysis, (2002), 26. When phenol is treated with CHCl 3 and NaOH, the, product formed is, (a) benzaldehyde, (b) salicylaldehyde, (c) salicylic acid, (d) benzoic acid. (2002), 27. Which of the following is correct?, (a) On reduction, any aldehyde gives secondary, alcohol., (b) Reaction of vegetable oil with H2SO4 gives, glycerine., (c) Alcoholic iodine with NaOH gives iodoform., (d) Sucrose on reaction with NaCl gives invert, sugar., (2001), 28. The correct acidic order of the following is, OH, OH, OH, , Alkalin e KMn O4, ,     , Z, , (2009), , 22. Ethylene oxide when treated with Grignard reagent, yields, (a) primary alcohol, (b) secondary alcohol, (c) tertiary alcohol, (d) cyclopropyl alcohol., (2006), 23. Which one of the following compounds is most, acidic?, , www.neetujee.com, , III., CH3, , (a) I > II > III, (c) II > III > I, , Anhyd. AlCl3, , (b) benzoic acid, (d) toluene., , II., , (2009), , 21. Consider the following reaction :, Z n d ust, C H Cl, Phenol  ,  X  3 , Y, , the product Z is, (a) benzaldehyde, (c) benzene, , (2005), , 24. Which one of the following will not form a yellow, precipitate on heating with an alkaline solution of, iodine?, (a) CH3 CH(OH)CH 3 (b) CH3 CH2 CH(OH)CH 3, (c) CH3OH, (d) CH3CH2OH (2004), , I., (c), , (c), , 29. Reaction of, formation of, , NO2, (b) III > I > II, (d) I > III > II, , (2001), , with RMgX leads to the, , (a) RCH2CH2OH, , (b) RCHOHCH3, , (c) RCHOHR, , (d), (1998), , 30. When 3,3-dimethyl-2-butanol is heated with H2SO4,, the major product obtained is, (a) 2,3-dimethyl-2-butene, (b) cis and trans isomers of 2,3-dimethyl-2-butene, (c) 2,3-dimethyl-1-butene, (d) 3,3-dimethyl-1-butene., (1995), , www.mediit.in

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Alcohols, Phenols and Ethers, , 89, , 59. The compound which does not react with sodium is, (a) CH 3 COOH, (b) CH3CHOHCH 3, (c) C2H5OH, (d) CH3OCH3, (1994), , 60. Which one is formed when sodium phenoxide is, heated with ethyl iodide?, (a) Phenetole, (b) Ethyl phenyl alcohol, (c) Phenol, (d) None of these (1988), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., , (a), (c), (b), (d), (c), (c), , 2., 12., 22., 32., 42., 52., , 3., 13., 23., 33., 43., 53., , (c), (b), (a), (a), (c), (b), , 4., 14., 24., 34., 44., 54., , (c), (d), (c), (a), (b), (a), , (c), (b), (c), (d), (a), (a), , 5., 15., 25., 35., 45., 55., , (d), (a), (c), (b), (d), (c), , 6., 16., 26., 36., 46., 56., , 7., 17., 27., 37., 47., 57., , (c), (b), (b), (a), (a), (a), , (d), (a), (c), (b), (a), (d), , 8., 18., 28., 38., 48., 58., , (c), (a), (b), (b), (a), (a), , 9., 19., 29., 39., 49., 59., , (c), (c), (a), (d), (b), (d), , 10., 20., 30., 40., 50., 60., , (d), (c), (a), (d), (c), (a), , Hints & Explanations, 1. (a) : All alcohols follow the general formula, CnH2n + 2O., CH3OH [CH2 + 2O] ; C2H5OH[C2H(2 × 2) + 2O], n = 1,, n=2, 2., , (c) :, CH3, , O, C, , CH3 + CH3MgCl, , Acetone, , CH3, , Methyl magnesium, chloride, , OMgCl, C CH3, CH3, , 6. (c) : As the compound is giving yellow precipitate, with NaOI that shows it is undergoing haloform reaction., Haloform reaction is shown by the compounds having, , HO, 2, , CH3, , OH, CH3, , C, , C, , or CH3, , O, CH3, , CH3, , CH, , group, , OH, , Hence, the compound A is, , tert-Butyl alcohol, , 3., , 2NaOH + I2, , (c) :, , NaOI + NaI + H2, , ‘Y’, , CH3, , CH3, , CH3, , CH, , aOI , N, , , CH3, , C O O H, , OH, O, , O2, , Cumene, , 4., , (c) :, , H+, H2 O, Cumene, hydroperoxide (A), , + CH3, Phenol, , CH3, , C, , CH3, , Acetone, , CH3, CHOH, , CH3, 2° alcohol, , Cu/573 K, , C O + H2, , 7., , (d) :, , CH3, Ketone, , 5. (d) : It is Reimer–Tiemann, reaction., The, electrophile formed is dichlorocarbene (:CCl2) which is, formed according to the following mechanism :, , www.neetujee.com, , www.mediit.in

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90, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , IV. (CH3)3C — CHOH — CH3, , CH3, CH, , CH3, , 13. (d) :, , +, , + CH3 CH, , 3, 3- Dimethyl-2-butanol, , CH3, Cumene, (P), , 14. (b) :, , 8. (c) : Electron withdrawing groups increase the, acidity while electron donating groups decrease the, acidity of phenol., 9., , (c) : This is Reimer–Tiemann reaction., , 10. (d) : 1° and 2° alcohols react with HCl in presence of, anhydrous ZnCl2 as catalyst while in case of 3° alcohols, ZnCl2 is not required., 11. (c) : The reaction is as follows :, Acid + NaHCO3  Sodium salt of acid + H2CO3, (soluble), , 15. (a) : Since, phenols and carboxylic acids are more, acidic than aliphatic alcohols, we find that cyclohexanol, (I) is least acidic. Out of the two given phenols, III is, more acidic than IV. This is because of the presence of, three highly electron withdrawing –NO2 groups on the, benzene ring which makes the O—H bond extremely, polarized. This facilitates the release of H as H+. Thus,, III > IV., , Among all the given compounds, o-nitrophenol is weaker, acid than HCO 3– . Hence, it does not react with NaHCO 3., , In acetic acid, the electron withdrawing, , 12. (b) : The iodoform test is positive for alcohols with, formula R — CHOH — CH3. Among C6H14O isomers,, the ones with positive iodoform test are:, I., CH3 — CH2 — CH2 — CH2 — CHOH — CH3, , in the —COOH group polarises the O—H bond and, increases the acidic strength. Acetic acid is therefore, more acidic than phenol or cyclohexanol.,  The order of acidic character is III > II > IV > I., , 2-Hexanol, , II. CH3 —CH2 — CH(CH3)— CHOH — CH3, 3- Methyl-2-pentanol, , III. (CH3)2CH — CH2 — CHOH — CH3, 4- Methyl-2-pentanol, , www.neetujee.com, , 16. (b) : Phenol is most acidic of all the given, compounds., In phenol, the electron withdrawing phenyl ring polarizes, the O—H bond, thereby facilitating the release of H as, H+ and hence, phenol is most acidic., , www.mediit.in

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91, , Alcohols, Phenols and Ethers, In, , OH, , , the electron withdrawing effect of, , phenyl ring is somewhat diminished by the — CH2 group, and it is therefore, less acidic than phenol. In (c) and (d),, — OH group is attached to alkyl groups which, due to, their +I effect reduce the polarity of — OH bond and so,, the acidic strength is low., 17. (a) : In phenols, the presence of electron releasing, groups decrease the acidity, whereas presence of electron, withdrawing groups increase the acidity, compared to, phenol. Among the meta and para-nitrophenols, the, later is more acidic as the presence of — NO2 group at, para position stabilises the phenoxide ion to a greater, extent than when it is present at meta position. Thus,, correct order of acidity is :, para-nitrophenol > meta-nitrophenol > phenol, (iv), (iii), (i), > methyl phenol, (ii), , Zn dust, , 21. (b) :, , + ZnO, Benzene(X), , Phenol, , COOH, , CH3, CH3 Cl anhy. AlCl3, Friedel-Cra s’, alkylation, , alk.KMnO4, Benzoic acid ( Z), , Toluene (Y), , 22. (a) :, H+, , CH3CH2CH2OMgBr  CH3CH2CH2OH, primary alcohol, , 23. (c) : Phenols are much more acidic than alcohols,, due to the stabilisation of phenoxide ion by resonance., , 18. (a) :, —NO2 is the electron withdrawing group and helps, in stabilizing the negative charge on the oxygen hence, equilibrium shifts in forward direction and more H+ ions, remove easily. Hence, it is most acidic., OH, , O, , CH3, , –, , CH3, , + H+, 19. (c) : C 2 H5OH PBr3,  C2 H5Br, Ethanol, , (X ), , alc.KOH, dehydrohalogenation, , , , 20. (c) : When 1,2-diol like ethylene glycol is treated, with HIO4, each alcoholic group is oxidised to a carbonyl, group by HIO4. Since in glycol, both the –OH groups are, terminal, so oxidation would yield two formaldehyde, molecules., www.neetujee.com, , —CH3 is the electron donating group. Hence, electron, density increases on the oxygen and destabilizes the, product. Thus, equilibrium shifts in backward direction., 24. (c) : Formation of a yellow precipitate on heating a, compound with an alkaline solution of iodine is known, as iodoform reaction. Methyl alcohol does not respond, to this test. Iodoform test is exhibited by ethyl alcohol,, acetaldehyde, acetone, methyl ketones and those alcohols, which possess CH3CH(OH)– group., 25. (c) : n-Propyl alcohol on oxidation with potassium, dichromate gives an aldehyde which on further oxidation, gives an acid. Both aldehyde and acid contain the same, number of C atoms as the original alcohol., i.e.,, CH3CH2CH2OH, www.mediit.in

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25C, 2, , 2 7, , CH CH CHO, 2, , 4, , KC,   , , r O /H SO, , 3, , 2, , K Cr O /H SO, , 2 2 7 2 4, ,    ,  CH3CH2COOH, , www.neetujee.com, , www.mediit.in

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92, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , Isopropyl alcohol on oxidation gives a ketone with the, same number of C atoms as the original alcohol., , 29. (a) :, , 26. (b) : This reaction is called Reimer—Tiemann, reaction., 30. (a) :, , CH3, CH3, , CH, , C, , +OH, , 2 CH3, , +, , CH3, , –H2 O, , CH3, , CH, , CH3, C, , CH3, , CH3, (2° carbocation), , 27. (c) : C2 H5 OH + 4I2 + NaOH, CHI3 + NaI + HCOONa + H2 O, Iodoform is a pale yellow solid which crystallises in, hexagonal plates., , CH3, CH, , 28. (b) : Phenol exists as a resonance hybrid of the, following structures., , +, , C, , C, , CH, , H, , CH3, , (3° carbocation), , 31. (d) :, 3, Thus, due to resonance the oxygen atom of the, — OH group acquires a positive charge and hence attracts, electron pair of the O – H bond leading to the release of, hydrogen atom as proton., OH, , O–, , 32. (a) :, , Mechanism:, +H, , Phenoxide ion, , Once the phenoxide ion is formed it stabilises itself by, resonance which is more stable than the parent phenol as, there is no charge separation., 33. (a), , Effect of substituent : Presence of electron withdrawing, groups (— NO2, — X, — CN) increase the acidity of, phenols while the presence of electron releasing groups, (— NH2, — CH3)decrease the acidity of phenols. This, explains the following order of acidity :, p-nitrophenol > phenol > p-cresol., , www.neetujee.com, , 34. (d) : — OCH3, — CH3 are electron donating groups, and decrease the acidic character of phenols. — NO2, is, electron withdrawing group and tends to increase the, acidic character. Electron donating effect of — OCH3, group (+R effect) is more than that of — CH3 group (+I, effect). Thus, the order is, p-methoxyphenol < p-methylphenol < p-nitrophenol., , www.mediit.in

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Alcohols, Phenols and Ethers, , 93, , 35. (b) : 2° alcohols on oxidation give ketones, 1°, alcohols form aldehydes., 36. (a) : Primary, alcohol, undergoes, catalytic, dehydrogenation to give aldehyde., 37. (b) : 4-isomers are possible for C5H11OH., (i) CH3CH2CH2CH2CH2OH, , OCH3, 46. (a) :, , OH, , +H, , I, , + CH3 I, , Anisole, , Phenol, , 47. (a) : In Ph O H, the lone pair of oxygen is in, conjugation with phenyl group so, it is least basic among, the given compounds and is most difficult to protonate., , (ii), , 48. (a) : Ethers are readily attacked by HI to give an, alkyl halide and alcohol. But when heated with excess of, HI, the product alcohol first formed reacts further with, HI to form the corresponding alkyl iodide., , (iii), , ROR + 2HI, , Heat, , RI + RI + H2O, , (excess), , 49. (b) : In case of phenyl methyl, , (iv), , ether,, , phenyl oxonium ion, 38. (b) :, , , generates, , 3° carbocation, , which is very stable intermediate, thus it will react more, rapidly with HBr., 39. (d) : Phenol on reaction with excess bromine water, gives 2,4,6-tribromophenol., 40. (d) : 2-Methylpropan-2-ol reacts rapidly with Lucas, reagent at room temperature., Cl, , Cl, , methyl, , is formed by, , protonation of ether. The O—CH3 bond is weaker than, O—C6H5 bond as O—C6H5 has partial double bond, character. Therefore, the attack by I– ion breaks O—CH3, bond to form CH3I., Step I :, , Step II :, 50. (c) : Williamson’s ether synthesis reaction involves, the treatment of sodium alkoxide with a suitable alkyl, halide to form an ether., 51. (c) : Williamson synthesis is the best method for, the preparation of ethers., , 41. (c) : –OH group being electron donor increases the, electron density in phenol. Thus, the electron density, in phenol is higher than that of toluene, benzene and, chlorobenzene., 42. (c) :, , 52. (b) : (a) CH3CHO + RMgX, , ether, , (b) C6 H5OH  NaOH ,  C6 H5ONa, H O, 2, , 43. (b) : Treatment of phenol with CHCl3 and aqueous, hydroxide introduces — CHO group, onto the aromatic, ring generally ortho to the — OH group. This reaction is, known as Reimer—Tiemann reaction., , Sodium, phenoxide, , (c), , 44. (a), ZnO-C r O, , 2 , 3 CH OH, 45. (d) : CO + 2H2   , 3, , www.neetujee.com, , www.mediit.in

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94, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 55. (c) : The alkyl iodide produced depends on the, nature of the alkyl groups. If one group is Me and the, other a primary or secondary alkyl group, it is methyl, iodide which is produced. This can be explained on the, assumption that the mechanism is SN2, and because of, the steric effect of the larger group, I– attacks the smaller, methyl group., When the substrate is a methyl t-alkyl ether, the, products are t-RI and MeOH. This can be explained by, SN1 mechanism, the carbonium ion produced being the, t-alkyl since tertiary carbonium ion is more stable than a, primary or secondary carbonium ion., 56. (a) : With cold HI, a mixture of alkyl iodide and, alcohol is formed. In the case of mixed ethers, the halogen, atom attaches to a smaller and less complex alkyl group., CH3OCH(CH3)2 + HI  CH3I + (CH3)2CHOH, , (d), , 53. (a) : CH3 CH2 CH, , CH 2, , HBr/H2O2, , 57. (d) :, The above reaction is called as Williamson’s synthesis., 58. (a) : Diethyl ether is a saturated compound, so it is, resistant to nucleophilic attack by a hydroxyl ion (OH–)., Other compounds have unsaturation and the unsaturated, ‘C’ atom bears partial +ve charge, therefore they undergo, easy nucleophilic attack by OH– ion., , 54. (a) :, , CH3, CH3, , C, , CH3, , 59. (d) : Ethers are very inert. The chemical inertness of, ethers is due to absence of active group in their molecules., Since CH3 – O – CH3 is inert and it does not contain, active group, therefore it does not react with sodium., , CH, , 60. (a) : Phenetole is formed when sodium phenoxide, is heated with ethyl iodide., , CH3, , H, O, CH 3, +, , CH3, , CH3, , +, , C + CH3OH, Methyl alcohol, I–, , 3, , CH3, , C, , , , I, , C6 H5ONa  C2 H5 I ,  C6 H5OC2 H5, Phenetole, , CH3, , , www.neetujee.com, , www.mediit.in

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96, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (c) Cross Cannizzaro’s reaction, (d) Cross Aldol condensation., , (NEET 2020), , 11. Consider the reactions,, Silver mirror, observed, , Identify A, X, Y and Z., (a) A-Methoxymethane, X-Ethanol,, Y-Ethanoic acid, Z-Semicarbazide., (b) A-Ethanal, X-Ethanol, Y-But-2-enal,, Z-Semicarbazone., (c) A-Ethanol, X-Acetaldehyde,, Y-Butanone, Z-Hydrazone., (d) A-Methoxymethane, X-Ethanoic acid,, Y-Acetate ion, Z-Hydrazine., (NEET 2017), 12. Of the following, which is the product formed, when cyclohexanone undergoes aldol condensation, followed by heating?, (b), , (a), , OH, , O, , O, (c), , (d), O, , O, , OH, (NEET 2017), , 13. The correct structure of the product ‘A’ formed in, the reaction, O, H2 gas, 1 atmosphere, , (d), , OH, , (d), (NEET-II 2016), , 14. Which of the following reagents would distinguish, cis-cyclopenta-1,2-diol from the trans-isomer?, (a) MnO 2, (b) Aluminium isopropoxide, (c) Acetone, (d) Ozone, (NEET-I 2016), , www.neetujee.com, , CHO, , CHO, , +, , (c), , COCH3, , (b), , (2014), , NO2, CH3, 19. The order of stability of the following tautomeric, compounds is, , OH, , OH, , CHO, , (a), , (c), , O, (b), , (a), , 17. Reaction of a carbonyl compound with one of the, following reagents involves nucleophilic addition, followed by elimination of water. The reagent is, (a) hydrazine in presence of feebly acidic solution, (b) hydrocyanic acid, (c) sodium hydrogen sulphite, (d) a Grignard reagent., (2015), 18. Which one is most reactive towards nucleophilic, addition reaction?, , A is, , Pd/carbon, ethanol, , OH, , 15. The correct statement regarding a carbonyl, compound with a hydrogen atom on its alphacarbon, is, (a) a carbonyl compound with a hydrogen atom, on its alpha-carbon rapidly equilibrates with its, corresponding enol and this process is known, as carbonylation, (b) a carbonyl compound with a hydrogen atom, on its alpha-carbon rapidly equilibrates with its, corresponding enol and this process is known, as keto-enol tautomerism, (c) a carbonyl compound with a hydrogen atom, on its alpha-carbon never equilibrates with its, corresponding enol, (d) a carbonyl compound with a hydrogen atom, on its alpha-carbon rapidly equilibrates with its, corresponding enol and this process is known, as aldehyde-ketone equilibration., (NEET-I 2016), 16. The product formed by the reaction of an aldehyde, with a primary amine is, (a) carboxylic acid, (b) aromatic acid, (c) Schiff ’s base, (d) ketone. (NEET-I 2016), , CH2, , C, , O, CH2, , C, , CH3, O, , (I), , CH3, , C, , O, CH2, , C, , CH3, , (II), , O, , OH, CH3, , C, , CH, , C, , CH3, , (III), , www.mediit.in

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Aldehydes, Ketones and Carboxylic Acids, (a) II > I > III, (c) I > II > III, , 97, , (b) II > III > I, (d) III > II > I, (NEET 2013), , 20. Predict the products in the given reaction., CHO, , 24. Which of the following compounds will give a, yellow precipitate with iodine and alkali?, (a) Acetophenone, (b) Methyl acetate, (c) Acetamide, (d) 2-Hydroxypropane, , 50% KOH, , Cl, , Cl, , OH, , COO–, OH, COO–, , CH2 OH, , (c), Cl, , 26. The order of reactivity of phenyl magnesium, bromide (PhMgBr) with the following compounds :, CH3, , CH3, C O, CH C, 3, H, , Cl, CH2 OH, , (d), , 25. Clemmensen reduction of a ketone is carried out in, the presence of which of the following?, (a) Glycol with KOH, (b) Zn-Hg with HCl, (c) LiAlH4, (d) H2 and Pt as catalyst, (2011), , Cl, CH2 OH, , (b), , (Mains 2012), , CH2 COO–, , CH2 OH, , (a), , OH, , I, , OH, OH, , (2012), , 21. Acetone is treated with excess of ethanol in the, presence of hydrochloric acid. The product obtained, is, O, (a) CH3CH2CH2 — C — CH3, , O, (b) CH3CH2CH2 — C — CH2CH2CH3, , (d) (CH3)2C, , OC2H5, , III, , (b) II > I > III, (d) I > II > III, (Mains 2011), , 27. Which of the following reactions will not result in, the formation of carbon-carbon bonds?, (a) Reimer–Tiemann reaction, (b) Cannizzaro reaction, (c) Wurtz reaction, (d) Friedel–Crafts acylation, (2010), , (b) H3C, , (a) CH3, , OH, O OH, , 23. Consider the reaction :, RCHO + NH2NH2  RCH=N — NH2, What sort of reaction is it?, (a) Electrophilic addition-elimination reaction, (b) Free radical addition-elimination reaction, , (c) CH3, , OH, OH, , (2012), , 22. CH3CHO and C 6H5 CH2 CHO can be distinguished, chemically by, (a) Benedict’s test, (b) iodoform test, (c) Tollens’ reagent test, (d) Fehling’s solution test., (2012), , www.neetujee.com, , Ph, CO, Ph, , II, , (a) III > II > I, (c) I > III > II, , OC2H5, OC2H5, , O and, , 28. Which one of the following compounds will be most, readily dehydrated?, O, O, , OH, (c) (CH3)2C, , (c) Electrophilic substitution-elimination reaction, (d) Nucleophilic addition-elimination reaction, (Mains 2012), , (d) CH3, , O, (Mains 2010), , 29. Following compounds are given,, (i) CH3CH2OH, (ii) CH3COCH3, (iii), , (iv) CH3OH, , Which of the above compound(s), on being warmed, with iodine solution and NaOH, will give iodoform?, (a) (i), (iii) and (iv), (b) Only (ii), (c) (i), (ii) and (iii), (d) (i) and (ii), (Mains 2010), , www.mediit.in

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98, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 30. Acetophenone when reacted with a base, C2H5ONa,, yields a stable compound which has the structure, CH3 CH3, C C, (a), OH OH, , 37. The major organic product formed from the, following reaction :, O, , (i) CH3NH2, , O NHCH3, , (a), , HNCH3, , HNCH3, , OH OH, , (c), C, , (c), , CH3, , CH C, O, , CH, , CH2 C, , CH3, , O, , HNCH3, , (d), , OH, , OH, , (d), , (2008), , 31. A strong base can abstract an -hydrogen from, (a) ketone, (b) alkane, (c) alkene, (d) amine., (2008), 32. Reduction of aldehydes and ketones into, hydrocarbons using zinc amalgam and conc. HCl is, called, (a) Cope reduction (b) Dow reduction, (c) Wolff–Kishner reduction, (d) Clemmensen reduction., (2007), 33. Which one of the following on treatment with 50%, aqueous sodium hydroxide yields the corresponding, alcohol and acid?, (a) C6H5CHO, (b) CH3CH2CH2CHO, O, , 3, , CH3 CH(OH)COOH, an asymmetric centre is generated. The acid obtained, would be, (a) D-isomer, (b) L-isomer, (c) 50% D + 50% L-isomer, (d) 20% D + 80% L-isomer., (2003), 39. When m-chlorobenzaldehyde is treated with 50%, KOH solution, the product(s) obtained is (are), COO–, CH2OH, (a), +, OH, (b), , 34. The product formed in aldol condensation is, (a) a beta-hydroxy aldehyde or a beta-hydroxy, ketone, (b) an alpha-hydroxy aldehyde or ketone, (c) an alpha, beta unsaturated ester, (d) a beta-hydroxy acid., (2007), 35. Nucleophilic addition reaction will be most favoured, in, (a) CH 3 CHO, O, (b) CH3 — CH2 — CH 2C— CH3, (c) (CH ) C O, (d) CH CH CHO (2006), , Cl, , Cl, OH, , CH— CH, , (c), Cl, , Cl, OH, , OH, , CH— CH, , (d), OH, , OH, , (2003), , 40. A and B in the following reactions are :, , , (a) A=RRC, , OH, COOH , B =NH3, , (b) A=RR C, , CN B =H O, ,, 3, OH, , 2, , 36. A carbonyl compound reacts with hydrogen cyanide, to form cyanohydrin which on hydrolysis forms a, racemic mixture of -hydroxy acid. The carbonyl, compound is, (a) formaldehyde, (b) acetaldehyde, (c) acetone, (d) diethyl ketone. (2006), , CH2OH, +, , OH, (2007), , 3, , OH, COO–, , (c) CH3 —C— CH3 (d) C6H5CH2CHO, , 3 2, , (2005), , 38. In this reaction :, HOH, , CH CHO + HCN  CH CH(OH)CN  , 3, , www.neetujee.com, , (b), , CH CH, , (b), , is, , (ii) LiAlH4 (iii) H2 O, , (c) A = RRCHCN, B = NaOH, (d) A=RRC, , CN B =LiAlH, ,, 4, OH, , (2003), , www.mediit.in

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102, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 78. In a set of the given reactions, acetic acid yielded a, product C., C6H6, , (a) sodium hydroxide, (c) calcium chloride, , CH3 COOH + PCl  A, 5, , B, , Anh. AlCl3, , C 2H5MgBr, ether, , Product C would be, (a) CH3CH(OH)C2H5 (b) CH3COC6H5, (c) CH3CH(OH)C6H5, C2H5, (d) CH3 – C(OH)C6H5, , C, , 86. The compound formed when malonic acid is heated, with urea is, (a) cinnamic acid, (b) butyric acid, (c) barbituric acid, (d) crotonic acid. (1989), (2003), , 79. Ethyl benzoate can be prepared from benzoic acid, by using, (a) ethyl alcohol, (b) ethyl alcohol and dry HCl, (c) ethyl chloride, (d) sodium ethoxide., (2000), 80. Reduction by LiAlH4 of hydrolysed product of an, ester gives, (a) two alcohols, (b) two aldehydes, (c) one acid and one alcohol, (d) two acids., (2000), 81. Which one of the following compounds will react, with NaHCO3 solution to give sodium salt and carbon, dioxide?, (a) Acetic acid, (b) n-Hexanol, (c) Phenol, (d) Both (b) and (c) (1999), 82. Which one of the following product is formed when, adipic acid is heated?, CH2CH2CO, CH2CH2COOH, (a), O (b), CH2CH2CO, CH2CH2COOH, (c), , (d), , CH2 — CH2, CH2 — CH2, CH2 — CH2, CH2 — CH2, , O, , C, , O, , (1995), , 83. An acyl halide is formed when PCl5 reacts with an, (a) amide, (b) ester, (c) acid, (d) alcohol., (1994), 84. Benzoic acid gives benzene on being heated with X, and phenol gives benzene on being heated with Y., Therefore, X and Y are respectively, (a) soda-lime and copper, (b) Zn dust and NaOH, (c) Zn dust and soda-lime, (d) soda-lime and zinc dust., (1992), 85. A is a lighter phenol and B is an aromatic carboxylic, acid. Separation of a mixture of A and B can be, carried out easily by using a solution of, , www.neetujee.com, , (b) sodium sulphate, (d) sodium bicarbonate., (1992), , 87. Among the following the strongest acid is, (a) CH3COOH, (b) CH2ClCH2COOH, (c) CH2ClCOOH, (d) CH3CH2COOH, (1988), 88. Which of the following is the correct decreasing, order of acidic strength of, (i) methanoic acid, (ii) ethanoic acid, (iii) propanoic acid, (iv) butanoic acid, (a) (i) > (ii) > (iii) > (iv) (b) (ii) > (iii) > (iv) > (i), (c) (i) > (iv) > (iii) > (ii) (d) (iv) > (i) > (iii) > (ii), (1988), , 12.A Derivatives of Carboxylic Acids, 89. Match the compounds given in List-I with List-II, and select the suitable option using the codes given, below., List-I, List-II, (A) Benzaldehyde, (i) Phenolphthalein, (B) Phthalic anhydride (ii) Benzoin, condensation, (C) Phenyl benzoate, (iii) Oil of wintergreen, (D) Methyl salicylate, (iv) Fries rearrangement, (a) (A)-(iv), (B)-(i), (C)-(iii), (D)-(ii), (b) (A)-(iv), (B)-(ii), (C)-(iii), (D)-(i), (c) (A)-(ii), (B)-(iii), (C)-(iv), (D)-(i), (d) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii) (Mains 2011), 90. Among the given compounds, the most susceptible, to nucleophilic attack at the carbonyl group is, (a) CH3COOCH3, (b) CH3CONH2, (c) CH3COOCOCH3 (d) CH3COCl, (2010), 91. The relative reactivities of acyl compounds towards, nucleophilic substitution are in the order of, (a) acid anhydride > amide > ester > acyl chloride, (b) acyl chloride > ester > acid anhydride > amide, (c) acyl chloride > acid anhydride > ester > amide, (d) ester > acyl chloride > amide > acid anhydride., (2008), 92. Self condensation of two moles of ethyl acetate in, presence of sodium ethoxide yields, (a) ethyl propionate, (b) ethyl butyrate, (c) acetoacetic ester, (d) methyl acetoacetate., (2006), , www.mediit.in

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Aldehydes, Ketones and Carboxylic Acids, , 103, , 93. Which one of the following esters cannot undergo, Claisen self-condensation?, (a) C6H5CH2COOC2H5 (b) C6H5COOC2H5, (c) CH3CH2CH2CH2COOC2H5, (d) C6H11CH2COOC2H5, (1998), , 94. Sodium formate on heating yields, (a) oxalic acid and H2, (b) sodium oxalate and H2, (c) CO2 and NaOH, (d) sodium oxalate., , (1993), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., 61., 71., 81., 91., , (c), (b), (d), (a), (a), (d), (c), (d), (a), (c), , 2., 12., 22., 32., 42., 52., 62., 72., 82., 92., , 3., 13., 23., 33., 43., 53., 63., 73., 83., 93., , (b), (a), (b), (d), (c), (d), (c), (d), (a), (c), , 4., 14., 24., 34., 44., 54., 64., 74., 84., 94., , (a), (b), (d), (a), (d), (b), (d), (c), (c), (b), , (a), (c), (a,d), (a), (a), (a), (c), (c), (d), (b), , 5., 15., 25., 35., 45., 55., 65., 75., 85., , (c), (b), (b), (a), (c), (b), (b), (d), (d), , 6., 16., 26., 36., 46., 56., 66., 76., 86., , (a), (c), (d), (b), (d), (d), (b), (a), (c), , 7., 17., 27., 37., 47., 57., 67., 77., 87., , (a), (a), (b), (b), (d), (a), (d), (b), (c), , 8., 18., 28., 38., 48., 58., 68., 78., 88., , (a), (d), (c), (c), (a), (d), (c), (d), (a), , 9., 19., 29., 39., 49., 59., 69., 79., 89., , (a), (d), (c), (b), (a), (a), (b), (b), (d), , 10., 20., 30., 40., 50., 60., 70., 80., 90., , (d), (c), (c), (d), (c), (b), (a), (a), (d), , Hints & Explanations, , 1., , Cl2/h, , (c) :, Toluene, , H2O, 373 K, , Benzal chloride, (X), , Benzaldehyde, , H2, Pd/BaSO4, , It is Rosenmund’s reduction., 4. (a) : Secondary alcohol on oxidation gives a ketone, containing the same number of carbon atoms., CH3, CH3, , CH3 – CHOH [O], , CH3 – C, , 2-Hydroxypropane, , O, , Acetone, , (c) : This is Rosenmund reduction., H2, , R — C — Cl Pd/BaSO RCHO, 4, (P), O, BaSO4 prevents the aldehyde from being reduced and, acts as a poison to the palladium catalyst in this reaction., 6. (a) : On passing a steam of ozone through a solution, of olefin in an organic solvent, an ozonide is obtained., , www.neetujee.com, , R2C, , O, , CHR, , O ozonide O, , 2. (b) : Clemmensen reduction in presence of Zn-Hg, and conc. HCl reduces aldehydes and ketones to —CH2, group but carboxylic acid group remains unaffected., 3. (a) : O C — Cl, H—C O, , 5., , R, C CHR + O3, R, , CHO, , CHCl2, , CH3, , The ozonide on reduction with Zn and acid or, H2/Ni gives aldehydes and/or ketones., , R C O CHR H2/Ni R CO + RCHO, 2, R, O, O, The nature of these products helps in locating the, position of the double bond in olefin., 7., , (a) : CH3 — C, , HCl, ether, , N + 2H  , CH3 — CH NH, , CH3 — CH, , (X), , H O, , NH 2  CH3 — CHO + NH3, , , (X), , (Y), , Y = Acetaldehyde, 8. (a) : A tertiary alcohol is difficult to oxidise. But, when it is treated with an acidic oxidising agent under, some conditions, it is oxidised to ketone and then to, acids. Both the ketone and acid contain the lesser number, of carbon atoms than the starting alcohol., 9. (a) : The oxidation of toluene (C6H5CH3) with, chromyl chloride (CrO2Cl2) in CCl4 or CS2 to give, , www.mediit.in

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106, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 31. (a) : The base (OH–) ion removes one of the, -hydrogen atom (which is some what acidic) from, aldehydes and ketones to form a carbanion or the enolate, ion. The acidity of -hydrogen is due to resonance, stabilization of enolate anion., 32. (d) : Aldehydes and ketones are converted to, alkanes when treated with zinc amalgam and conc. HCl., This is known as Clemmensen reduction. Here C O, group is reduced to CH2 group., , 38. (c) : Lactic acid (CH 3 CH(OH)COOH) is an, optically active compound due to the presence of, asymmetric carbon atom. It exists in D- and L-form,, the ratio of which is found to be (1 : 1), i.e., a racemic, mixture is obtained., CHO, COOK, CH2 OH, 39. (b) :, , +, , KOH, , Cl, , Cl, , Cl, , m-chlorobenzaldehyde, , 33. (a) : Aldehydes which do not have -H atom,, in presence of 50% NaOH or 50% KOH undergo, disproportionation reaction to produce alcohol and, sodium salt of acid. This reaction is known as Cannizzaro, reaction. C6H 5CHO containing no -H atom undergoes, Cannizzaro reaction to produce benzyl alcohol and, sodium benzoate., , +, , or, , Cl, Cl, The above reaction is known as Cannizzaro’s reaction., CN, , 40. (d) : R — C — R HCN/KCN R — C — R, , C6 H5 CHO 50% NaOH C6 H5 CH2 OH + C6 H5 COONa, , 34. (a) : The aldehydes or ketones containing -H atom, in presence of dilute alkali undergo self condensation, reaction to form -hydroxyaldehyde or -hydroxyketone., This reaction is known as aldol condensation., 35. (a) : The reactivity of the carbonyl group towards, the addition reactions depends upon the magnitude, of the positive charge on the carbonyl carbon atom., The introduction of group with –I effect increases the, reactivity while introduction of alkyl group (+I effect), decreases the reactivity., O, O, O, CH3 — C — H > CH3CH2 — C — H > CH3 — C — CH3, O, > CH3CH2CH2 — C — CH3, +I-effect increases, reactivity decreases., H, , CH3—C—OH, , 36. (b) : CH3CHO+ HCN, , CN, CH3 —C— OH, COOH, O, + NH –CH, 2, , C, , NH – CH3, CH, , LiAlH 4, H2O, , CH2 NH2, , LiAlH4, , R —C— R, , (B), , , , OH, 41. (a) : They are resonating forms because the position, of the atomic nuclei remains the same and only electron, redistribution has occurred., –, , C — CH, , CH, , CH2 — C — CH3, O, , 2, , 3, , O–, , 42. (c) : Tollens’ reagent is a solution of ammoniacal, silver nitrate and used for the detection of — CHO, group. Aldehydes reduce Tollens’ reagent and itself gets, oxidised to convert Ag+ ions to Ag powder which forms, the silver coloured mirror in the test tube. So, this test is, also known as silver mirror test., R — CHO + [Ag(NH3)2]+  R — COO– + Ag, 43. (d) : O-atom is more electronegative than C-atom,, therefore O-atom bears partial –ve charge and C-atom to, which it is attached bear partial +ve charge., +, , CH2, , N–CH3, , 3, , OH, (A), , Lactic acid, , 37. (b) :, , www.neetujee.com, , O, , (Powder), , H, H2O, , COO–, , CH2 OH, , CH2 – CH, , CH– C O, H, , –, , C –O, H, , 44. (a) : It is a simple condensation reaction which, proceeds with elimination of water., R – CH, , O + H2 N– NH2 –H O RCH, 2, , N – NH2, , www.mediit.in

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Amines, , 113, , 29. The compound obtained by heating a mixture, of ethylamine and chloroform with ethanolic, potassium hydroxide (KOH) is, (a) an amide, (b) an amide and nitro compound, (c) an ethyl isocyanide, (d) an alkyl halide., (1997), 30. An aniline on nitration gives, CH3, NH2, (a), , CN, , CONH 2, , (c), , (d), (NEET-II 2016), , (b), , NO2, , NO2, NH2, (c), , an unstable compound ‘B’. ‘B’, on treatment with, phenol, forms a beautiful coloured compound, ‘C’ with the molecular formula C12H10N2O. The, structure of compound ‘A’ is, NO2, NH2, (a), (b), , NO2, , 36. In the following reaction, the product (A) is, +, , N, , NO2, , NCl, , –, , NH2, H+, , +, , (d) both (a) and (c)., , (A), , Yellow dye, , (1996), 31. The action of nitrous acid on an aliphatic primary, amine gives, (a) secondary amine, (b) nitro alkane, (c) alcohol, (d) alkyl nitrite. (1994), 32. Which one of the following order is wrong, with, respect to the property indicated?, (a) Benzoic acid > phenol > cyclohexanol (acid, strength), (b) Aniline > cyclohexylamine > benzamide (basic, strength), (c) Formic acid > acetic acid > propanoic acid (acid, strength), (d) Fluoroacetic acid > chloroacetic acid >, bromoacetic acid (acid strength), (1994), 33. For carbylamine reaction, we need hot alcoholic, KOH and, (a) any primary amine and chloroform, (b) chloroform and silver powder, (c) a primary amine and an alkyl halide, (d) a monoalkylamine and trichloromethane., (1992), , 13.7 Methods of Preparation of Diazonium, Salts, 34. Which of the following will be most stable diazonium, salt RN2+X –?, + –, , + –, , (a) CH3N2 X, (c) CH3CH N +X–, 2, , (b) C6H5N2 X, (d) C H CH N +X– (2014), , 2, , 6, , 5, , N, , N NH, NH2, , (b), , N, , N, NH2, , (c), , N, , N, , (d), , N, , N, , NH2, , 35. A given nitrogen-containing aromatic compound, ‘A’ reacts with Sn/HCl, followed by HNO2 to give, , (2014), , 37. In the reaction, , A is, (a) H3PO 2 and H 2O, (c) HgSO4/H2SO 4, , (b) H+/H2O, (d) Cu2Cl2 (NEET 2013), , 38. Aniline in a set of the following reactions yielded a, coloured product Y., , NH2, NaNO2/HCl, (273-278 K), , X, , N, N-dimethylaniline, , Y, , The structure of ‘Y’ would be, (a), , N, , N, , N, , CH3, CH3, , 2 2, , 13.9 Chemical Reactions, , www.neetujee.com, , (a), , CH3, (b) HN, , CH3, NH, , NH, , www.mediit.in

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114, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , (c) H3C, , N, , (b) 1, 2, 4-Trinitrobenzene, (c) 1, 2-Dinitrobenzene, (d) 1, 3-Dinitrobenzene, , NH2, , N, , CH3, , CH3, NN, , (d) HN, , NH, (2010, 2008, 2004), , 43. What is the product obtained in the following, reaction?, NO2, Zn, , 39. Aniline in a set of reactions yielded a product D., NH2, , NaNO 2, HCl, , A, , CuCN, , (NEET 2013), , ?, , NH4Cl, , NHOH, , B, H2, Ni, , C, , (a), , HNO2, , D, , N, , The structure of the product D would be, (a) C6H5NHOH, (b) C6H5NHCH2CH3, (c) C6H5CH2NH2, (d) C6H5CH2OH (2005), 40. Aniline is reacted with bromine water and the, resulting product is treated with an aqueous solution, of sodium nitrite in presence of dilute hydrochloric, acid. The compound so formed is converted into a, tetrafluoroborate which is subsequently heated to, dry. The final product is, (a) p-bromoaniline, (b) p-bromofluorobenzene, (c) 1, 3, 5-tribromobenzene, , (b), , O–, , , , , N, +, , NH2, , (d), , (2011), , CN, , 44., , , , H3 O, , , P, , + CH3 MgBr, OCH 3, , 13.A Other Nitrogen Containing Compounds, , Product ‘P’ in the above reaction is, , 41. Which one of the following nitro-compounds does, not react with nitrous acid?, H2, C, H2, H3 C, (b), CH, NO2, (a) H3C C, NO2, H3C, C, H2, CH3, H3 C, C, H, C, C, NO, H, C, (d) 3, 2, (c) 3, H3 C, H NO2, O, (NEET-II 2016), 42. Nitrobenzene on reaction with conc. HNO3/H2SO4, at 80–100°C forms which one of the following, products?, (a) 1, 4-Dinitrobenzene, , N, , (c), , (1998), , (d) 2, 4, 6-tribromofluorobenzene., , N, , OH, , O, , CH– CH3, , C– CH3, , (a), , (b), OCH3, , OCH3, CHO, , COOH, , (c), , (d), , (2002), , OCH3, , OCH3, , 45. Which product is formed, when acetonitrile is, hydrolysed partially with cold concentrated HCl?, (a) Methyl cyanide, (b) Acetic anhydride, (c) Acetic acid, (d) Acetamide, (1995), , ANSWER KEY, , 1., 11., 21., 31., 41., , (a), (c), (b), (c), (c), , www.neetujee.com, , 2., 12., 22., 32., 42., , (c), (a), (c), (b), (d), , 3., 13., 23., 33., 43., , (c), (b), (c), (a), (a), , 4., 14., 24., 34., 44., , (c), (a), (d), (b), (b), , 5., 15., 25., 35., 45., , (a), (d), (a), (b), (d), , 6., 16., 26., 36., , (a), (c), (c), (d), , 7., 17., 27., 37., , (a), (c), (b), (a), , 8., 18., 28., 38., , (c), (a), (b), (a), , 9., 19., 29., 39., , (a), (a), (c), (d), , 10., 20., 30., 40., , (a), (a), (d), (d), , www.mediit.in

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, , , , , , , , , CHAPTER, , 14, , Biomolecules, , 14.1 Carbohydrates, 1., , 2., , 3., , 4., , 5., , Sucrose on hydrolysis gives, (a) -D-glucose + -D-fructose, (b) -D-glucose + -D-glucose, (c) -D-glucose + -D-fructose, (d) -D-fructose + -D-fructose., , CH, , CH, , H C OH, , HO, , (b), , HO C H, H C OH, , (NEET 2020), , CHO, CHO, HO, H, H, OH, H, HO, H, HO, CH2OH, CH2OH, respectively, is, (a) L-erythrose, L-threose, L-erythrose, D-threose, (b) D-threose, D-erythrose, L-threose, L-erythrose, (c) L-erythrose, L-threose, D-erythrose, D-threose, (d) D-erythrose, D-threose, L-erythrose, L-threose., (NEET-II 2016), Which one given below is a non-reducing sugar?, (a) Glucose, (b) Sucrose, (c) Maltose, (d) Lactose (NEET-I 2016), D(+)-glucose reacts with hydroxyl amine and yields, an oxime. The structure of the oxime would be, , H C OH, , CH2OH, , CH2OH, , CH, , CH, , NOH, , C OH, CH2OH, , NOH, , H C OH, HO C H, , H C OH, , (c) HO C H, , 6., , C H, , H C OH, , HO C H, , H, , NOH, , HO C H, , HO C H, , (a), , The difference between amylose and amylopectin is, (a) amylopectin have 1  4 -linkage and 1  6, -linkage, (b) amylose have 1  4 -linkage and 1  6, -linkage, (c), amylopectin have 1  4 -linkage and 1  6, -linkage, (d), amylose is made up of glucose and galactose., (NEET 2018), The correct corresponding order of names of four, aldoses with configuration given below, CHO, CHO, OH, H, H, HO, H, OH, OH, H, CH2OH, CH2OH, , www.neetujee.com, , NOH, , (d), , H C OH, H C OH, CH2 OH, , (2014), , Which one of the following sets of monosaccharides, forms sucrose?, (a) -D-galactopyranose and -D-glucopyranose, (b) -D-glucopyranose and -D-fructofuranose, (c) -D-glucopyranose and -D-fructofuranose, (d) -D-glucopyranose and -D-fructopyranose, (2012), , 7., , Which one of the following statements is not true, regarding (+)–lactose?, (a) On hydrolysis (+)–lactose gives equal amount, of D(+)–glucose and D(+)–galactose., (b) (+)–Lactose is a -glucoside formed by the, union of a molecule of D(+)–glucose and a, molecule of D(+)–galactose., (c) (+)–Lactose is a reducing sugar and does not, exhibit mutarotation., (d) (+)–Lactose, C12H22O11 contains 8 –OH groups., (2011), , 8., , Which one of the following does not exhibit the, phenomenon of mutarotation?, (a) (+)–Sucrose, (b) (+)–Lactose, (c) (+)–Maltose, (d) (–)–Fructose (2010), , www.mediit.in

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120, , 9., , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , Fructose reduces Tollens’ reagent due to, (a) asymmetric carbons, (b) primary alcoholic group, (c) secondary alcoholic group, (d) enolisation of fructose followed by conversion, to aldehyde by base., (Mains 2010), , 10. Number of chiral carbons in -D-(+) glucose is, (a) five, (b) six, (c) three, (d) four., (2004), 11. Glycolysis is, (a) oxidation of glucose to glutamate, (b) conversion of pyruvate to citrate, (c) oxidation of glucose to pyruvate, (d) conversion of glucose to haem., , (2003), , 12. Cellulose is polymer of, (a) glucose, (b) fructose, (c) ribose, (d) sucrose., , (2002), , 13. Which of the following gives positive Fehling, solution test?, (a) Sucrose, (b) Glucose, (c) Fats, (d) Protein, (2001), 14. -D-glucose and -D-glucose are, (a) epimers, (b) anomers, (c) enantiomers, (d) diastereomers. (2000), 15. Which of the following is the sweetest sugar?, (a) Fructose, (b) Glucose, (c) Sucrose, (d) Maltose, (1999), 16. Glucose molecule reacts with X number of molecules, of phenyl hydrazine to yield osazone. The value of X, is, (a) two, (b) one, (c) four, (d) three., (1998), 17. The oxidation of glucose is one of the most important, reactions in a living cell. What is the number of ATP, molecules generated in cells from one molecule of, glucose?, (a) 28, (b) 38, (c) 12, (d) 18, (1995), 18. The -D-glucose and -D-glucose differ from each, other due to difference in carbon atom with respect, to its, (a) number of OH groups, (b) size of hemiacetal ring, (c) conformation, (d) configuration., (1995), 19. Chemically considering digestion is basically, (a) anabolism, (b) hydrogenation, (c) hydrolysis, (d) dehydrogenation., , www.neetujee.com, , 20. On hydrolysis of starch, we finally get, (a) glucose, (b) fructose, (c) both (a) and (b), (d) sucrose., , (1991), , 14.2 Proteins, 21. Which of the following is a basic amino acid?, (a) Serine, (b) Alanine, (c) Tyrosine, (d) Lysine (NEET 2020), 22. The non-essential amino acid among the following, is, (a) lysine, (b) valine, (c) leucine, (d) alanine., (NEET 2019), 23. Which structure(s) of proteins remain(s) intact, during denaturation process?, (a) Both secondary and tertiary structures, (b) Primary structure only, (c) Secondary structure only, (d) Tertiary structure only, (Odisha NEET 2019), 24. Which of the following compounds can form a, zwitter ion?, (a) Aniline, (b) Acetanilide, (c) Benzoic acid, (d) Glycine, (NEET 2018), 25. In a protein molecule various amino acids are linked, together by, (a) peptide bond, (b) dative bond, (c) -glycosidic bond, (d) -glycosidic bond., (NEET-I 2016), 26. Which of the statements about “Denaturation” given, below are correct?, (1) Denaturation of proteins causes loss of, secondary and tertiary structures of the protein., (2) Denaturation leads to the conversion of double, strand of DNA into single strand., (3) Denaturation affects primary structure which, gets distorted., (a) (2) and (3), (b) (1) and (3), (c) (1) and (2), (d) (1), (2) and (3), (Mains 2011), 27. Which functional group participates in disulphide, bond formation in proteins?, (a) Thioester, (b) Thioether, (c) Thiol, (d) Thiolactone (2005), 28. Which of the following structures represents the, peptide chain?, H, , O, , (a) – N – C – N – C – NH– C – NH–, (1994), , OH, , www.mediit.in

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Biomolecules, H, , 121, H, , (b) – N – C – C – C – C – N – C – C – C –, O, H, H, H, O, –, N, –, C, –, C, –, N, –, C, –, C, –, N, –, C, –, C, –, (c), O, O, H, H, O, –, N, –, C, –, C, –, C, –, N, –, C, –, C, –, N– C– C–C–, (d), O, H, , (2004), , 29. The correct statement in respect of protein, haemoglobin is that it, (a) functions as a catalyst for biological reactions, (b) maintains blood sugar level, (c) acts as an oxygen carrier in the blood, (d) forms antibodies and offers resistance to, diseases., (2004), 30. The helical structure of protein is stabilised by, (a) dipeptide bonds, (b) hydrogen bonds, (c) ether bonds, (d) peptide bonds., (2004), 31. Which is not true statement?, (a) -Carbon of -amino acid is asymmetric., (b) All proteins are found in L-form., (c) Human body can synthesise all proteins they, need., (d) At pH = 7 both amino and carboxylic groups, exist in ionised form., (2002), O, , .., 32. – C – NH –(peptide bond)., Which statement is incorrect about peptide bond?, (a) C – N bond length in proteins is longer than, usual bond length of N – C bond., (b) Spectroscopic analysis shows planar structure, of – C – NH – group., , O, (c) C – N bond length in proteins is smaller than, usual bond length of C – N bond., (d) None of the above., (2001), 33. Which is the correct statement?, (a) Starch is a polymer of -glucose., (b) Amylose is a component of cellulose., (c) Proteins are composed of only one type of, amino acid., (d) In cyclic structure of fructose, there are four, carbons and one oxygen atom., (2001), 34. Haemoglobin is, (a) a vitamin, (c) an enzyme, , www.neetujee.com, , (b) a carbohydrate, (d) a globular protein., (1997), , 35. The secondary structure of a protein refers to, (a) regular folding patterns of continuous portions, of the polypeptide chain, (b) three-dimensional structure, specially the bond, between amino acid residues that are distant, from each other in the polypeptide chain, (c) mainly denatured proteins and structures of, prosthetic groups, (d) linear sequence of amino acid residues in the, polypeptide chain., (1995), , 14.3 Enzymes, 36. During the process of digestion, the proteins present, in food materials are hydrolysed to amino acids. The, two enzymes involved in the process, enzyme (A), , proteins     polypeptides, enzyme (B ), ,    amino acids,, , are respectively, (a) invertase and zymase, (b) amylase and maltase, (c) diastase and lipase, (d) pepsin and trypsin., 37. Enzymes are made up of, (a) edible proteins, (b) proteins with specific structure, (c) nitrogen containing carbohydrates, (d) carbohydrates., , (2006), , (2002), , 38. Which of the following is correct?, (a) Cycloheptane is an aromatic compound., (b) Diastase is an enzyme., (c) Acetophenone is an ether., (d) All of these., (2001), 39. The function of enzymes in the living system is to, (a) catalyse biochemical reactions, (b) provide energy, (c) transport oxygen, (d) provide immunity., (1997), 40. Which of the following statements about enzymes is, true?, (a) Enzymes catalyse chemical reactions by, increasing the activation energy., (b) Enzymes are highly specific both in binding, chiral substrates and in catalysing their, reactions., (c) Enzymes lack in nucleophilic groups., (d) Pepsin is proteolytic enzyme., (1995), , www.mediit.in

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122, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 41. Enzymes take part in a reaction and, (a) decrease the rate of a chemical reaction, (b) increase the rate of a chemical reaction, (c) both (a) and (b), (d) none of these., (1993), , (a), (b), (c), (d), , 14.4 Vitamins, 42. Deficiency of vitamin B1 causes the disease, (a) convulsions, (b) beri-beri, (c) cheilosis, (d) sterility., (2012), 43. Which of the following is not a fat soluble vitamin?, (a) Vitamin B complex, (b) Vitamin D, (c) Vitamin E, (d) Vitamin A, (Mains 2011), 44. Which of the following vitamins is water soluble?, (a) Vitamin E, (b) Vitamin K, (c) Vitamin A, (d) Vitamin B, (2007), 45. The human body does not produce, (a) enzymes, (b) DNA, (c) vitamins, (d) hormones., , (2006), , 46. Vitamin B12 contains, (a) Fe (II), (c) Zn (II), , (2003), , (b) Co (III), (d) Ca (II), , phosphate linkage, H-bonding, glycosidic linkage, peptide linkage., , (Karnataka NEET 2013), , 50. The segment of DNA which acts as the instrumental, manual for the synthesis of the protein is, (a) ribose, (b) gene, (c) nucleoside, (d) nucleotide., (2009), 51. In DNA, the complimentary bases are, (a) adenine and guanine; thymine and cytosine, (b) uracil and adenine; cytosine and guanine, (c) adenine and thymine; guanine and cytosine, (d) adenine and thymine; guanine and uracil., (2008, 1998), 52. RNA and DNA are chiral molecules, their chirality, is due to, (a) chiral bases, (b) chiral phosphate ester units, (c) D-sugar component, (d) L-sugar component., (2007), 53. A sequence of how many nucleotides in messenger, RNA makes a codon for an amino acid?, (a) Three, (b) Four, (c) One, (d) Two, (2004), , 54. Chargaff’s rule states that in an organism, (a) amount of adenine (A) is equal to that of thymine, 14.5 Nucleic Acids, (T) and the amount of guanine (G) is equal to, that of cytosine (C), 47. The central dogma of molecular genetics states that, (b), amount of adenine (A) is equal to that of guanine, the genetic information flows from, (G) and the amount of thymine (T) is equal to, (a) Amino acids  Proteins  DNA, that of cytosine (C), (b) DNA  Carbohydrates  Proteins, (c), amount of adenine (A) is equal to that of cytosine, (c) DNA  RNA  Proteins, (C) and the amount of thymine (T) is equal to that, (d) DNA  RNA  Carbohydrates, of guanine (G), (NEET-II 2016), (d) amounts of all bases are equal., (2003), 48. The correct statement regarding RNA and DNA,, 55. Which of the following is correct about H-bonding, respectively is, in nucleotide?, (a) the sugar component in RNA is a arabinose and, (a) A – T, G – C, (b) A – G , T – C, the sugar component in DNA is ribose, (c) G – T, A – C, (d) A – A, T – T, (b) the sugar component in RNA is 2-deoxyribose, (2001), and the sugar component in DNA is arabinose, 56. An example of biopolymer is, (c) the sugar component in RNA is arabinose and, (a) teflon, (b) neoprene, the sugar component in DNA is 2-deoxyribose, (c) nylon-6, 6, (d) DNA., (1994), (d) the sugar component in RNA is ribose and the, 57. The couplings between base units of DNA is through, sugar component in DNA is 2-deoxyribose., (a) hydrogen bonding, (NEET-I 2016), (b) electrostatic bonding, 49. In DNA, the linkages between different nitrogenous, bases are, , www.neetujee.com, , (c) covalent bonding, (d) van der Waals’ forces., , (1992), , www.mediit.in

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Biomolecules, , 123, , 14.6 Hormones, 58. Which of the following statements is not correct?, (a) Ovalbumin is a simple food reserve in eggwhite., (b) Blood proteins thrombin and fibrinogen are, involved in blood clotting., (c) Denaturation makes the proteins more active., (d) Insulin maintains sugar level in the blood of a, human body., (NEET 2017), 59. Which of the following hormones is produced, under the conditions of stress which stimulate, glycogenolysis in the liver of human beings?, (a) Thyroxin, (b) Insulin, (c) Adrenaline, (d) Estradiol, (2014), 60. Which of the following hormones contains iodine?, (a) Testosterone, (b) Adrenaline, (c) Thyroxine, (d) Insulin, (2009), 61. Which of the following is an amine hormone?, (a) Insulin, (b) Progesterone, (c) Thyroxine, (d) Oxypurin (2008), 62. Which one of the following is a peptide hormone?, (a) Adrenaline, (b) Glucagon, (c) Testosterone, (d) Thyroxine (2006), , 63. The hormone that helps in the conversion of glucose, to glycogen is, (a) cortisone, (b) bile acids, (c) adrenaline, (d) insulin., (2004), 64. Which one is responsible for production of energy, in biochemical reaction?, (a) Thyroxine, (b) Adrenaline, (c) Oestrogen, (d) Progesterone (2000), , 14.A Lipids, 65. The cell membranes are mainly composed of, (a) fats, (b) proteins, (c) phospholipids, (d) carbohydrates., (2005), 66. Phospholipids are esters of glycerol with, (a) three carboxylic acid residues, (b) two carboxylic acid residues and one phosphate, group, (c) one carboxylic acid residue and two phosphate, groups, (d) three phosphate groups., (2003), 67. The number of molecules of ATP produced in the, lipid metabolism of a molecule of palmitic acid is, (a) 56, (b) 36, (c) 130, (d) 86, (1998), , ANSWER KEY, , 1., 11., 21., 31., 41., 51., 61., , (c), (c), (d), (b), (b), (c), (c), , 2., 12., 22., 32., 42., 52., 62., , (a), (a), (d), (a), (b), (c), (b), , 3., 13., 23., 33., 43., 53., 63., , (d), (b), (b), (a), (a), (a), (d), , 4., 14., 24., 34., 44., 54., 64., , (b), (b), (d), (d), (d), (a), (a), , 5., 15., 25., 35., 45., 55., 65., , (d), (a), (a), (a), (c), (a), (c), , 6., 16., 26., 36., 46., 56., 66., , (b), (d), (c), (d), (b), (d), (b), , 7., 17., 27., 37., 47., 57., 67., , (c), (b), (c), (b), (c), (a), (c), , 8., 18., 28., 38., 48., 58., , (a), (d), (c), (b), (d), (c), , 9., 19., 29., 39., 49., 59., , (d), (c), (c), (a), (b), (c), , 10., 20., 30., 40., 50., 60., , (d), (a), (b), (b), (b), (c), , Hints & Explanations, 1. (c) : In sucrose, two monosaccharides are held, together by a glycosidic linkage between C-1 of, -D-glucose and C-2 of -D-fructose., Hydrolysis, Sucrose, -D-glucose + -D-fructose, 2. (a) : Amylose is a linear polymer of -D-glucose, held by C1-C4 glycosidic linkage whereas amylopectin is, branched chain polymer of -D-glucose units in which, chain is held by C1-C4 glycosidic linkage while branching, occurs by C1-C6 glycosidic linkage., , www.neetujee.com, , 3., , (d) :, , CHO, CHO, H, H, OH HO, H, OH, H, OH, CH2 OH, CH2 OH, D-erythrose, , D-threose, , CHO, HO, H, HO, H, CH2 OH, L-erythrose, , CHO, H, OH, HO, H, CH2 OH, L-threose, , www.mediit.in

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124, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 4. (b) : All monosaccharides whether aldoses or, ketoses are reducing sugars. Disaccharides such as, sucrose in which the two monosaccharide units are, linked through their reducing centres i.e., aldehydic or, ketonic groups are non-reducing., 5., , (d) :, , CH, H C OH, , H C OH, HO C H, , HO C H, H, , C OH, , H, , C OH, , NOH, , +N HO2H, , CH2OH, D(+)-Glucose, , H, , C OH, , H, , C OH, CH2OH, Glucoxime, , 6. (b) : Sucrose is formed by the condensation of, -D-glucopyranose and -D-fructofuranose., 7. (c) : (+)-Lactose is a reducing sugar and all reducing, sugars show mutarotation., 8. (a) : Sucrose does not show mutarotation., Mutarotation is the phenomenon of change in optical, rotation shown by freshly prepared solutions of, sugars. However, this property is not exhibited by all, sugars. Only those sugars which have a free aldehyde, (—CHO) or ketone (>C O) group are capable of, showing mutarotation. Sucrose lacks free aldehyde or, ketone group and is therefore, incapable of showing, mutarotation., 9. (d) : Under alkaline conditions of the reagent,, fructose gets converted into a mixture of glucose and, mannose (Lobry de Bruyn van Ekenstein rearrangement), both of which contain the –CHO group and hence,, reduce Tollens’ reagent to give silver mirror test., 10. (d) : HO, H, C, , linkage between C1 of one glucose unit and C4 of the next, glucose unit., 13. (b) : Glucose reduces Fehling solution because, glucose has free –CHO group which is readily oxidised., 14. (b) : Glucose forms a stable hemiacetal between the, —CHO group and the —OH group on the 5 th carbon., In this process, the 1st ‘C’ atom becomes asymmetric, giving two isomers which differ in the configuration of, the asymmetric carbon. These two isomers are called as, anomers., CHO, OH, H, OH, OH, , H, HO, H, H, , * CHOH, , 2, 3, , H, , CH2OH, , Anomers, , 15. (a) :, , Fructose is the sweetest among all the sugars, and is highly soluble in water., 16. (d) : Glucose first reacts with phenyl hydrazine, giving phenylhydrazone. Then the adjacent —CHOH, group is oxidized by a 2nd phenyl hydrazine molecule, and itself is reduced to aniline. The resulting carbonyl, group reacts with 3rd phenyl hydrazine molecule giving, osazone., CHO, CH – OH, , + C H – NH – NH, 65, , (CH – OH)3, , – H2 O, 2, , CH2 – OH, (Glucose), , CH, , N, , NH, , C 6H5, , CH OH C H – NH–NH, 65, 2, (CH OH), , – (C6H5NH2, NH3), , CH, CO, , N NH C6H5, , (CH OH), 3, , CH2 OH, CH, C, , * CHOH, , N – NH – C6H5, N – NH – C6H5, , (CH – OH) 3, , * CH, , CH2 – OH, , CH2OH, , www.neetujee.com, , -D-glucose, , -D-glucose, , O, , 11. (c) : Glycolysis is the first stage in the oxidation, of glucose. It is an anaerobic process and involves the, degradation of glucose into two molecules of pyruvate, with the generation of two molecules of ATP., 12. (a) : Cellulose is a straight chain polysaccharide, composed of -D-glucose units joined by -glycosidic, , H, OH, , 4, 5, 6, , CH2 OH, , This structure of -D-glucose has four asymmetric, carbon atoms., , 1, 2, , HO, H, , OH, , OH, O, H O + HO 3, H, H 4, OH, OH, 5, H, 6, CH2OH, CH2OH, , HO, H, , D-glucose, , 3, , * CHOH, , 1, , H, H, , (Glucosazone), , 17. (b) : C6H12O6 + 6O2  6CO2 + 6H2O + 38ATP, , 18. (d) :, , H, H, HO, H, H, , 1, 2, 3, 4, 5, 6, , OH, OH, O, H, OH, , HO, H, HO, H, H, , 1, 2, 3, 4, 5, 6, , H, OH, O, H, OH, , CH2 OH, , CH2 OH, , -D-glucose, , -D-glucose, , www.mediit.in

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Biomolecules, , 125, , These isomers differ only in the orientation (or, configuration) at C1 atom., 19. (c), 20. (a) : Glucose is produced commercially by the, hydrolysis of starch by boiling it with dil. H2SO4 at 393 K, under pressure of 2-3 bar., 393 K, 2-3 bar, , 21. (d) : Lysine is a basic amino acid., COOH, H2N (CH2)4 C NH2, H, 22. (d), 23. (b) : During denaturation of proteins, 2° and 3°, structures are destroyed but 1° structure, remains intact., +, –, 24. (d) : HOOC — CH — NH, 2, Glycine, , 2, , OOC — CH2 — NH3, Zwitter ion, , 25. (a), 26. (c) : Denaturation does not change the primary, structure of protein., 27. (c) : Disulphide bond may be reduced to thiol by, means of reagents i.e., NaBH4, which shows the presence, of thiol group in disulphide bond formation., 28. (c) : In peptide linkage i.e., – CONH – group, the, carboxyl group of one amino acid molecule forms an, amide by combination with the amino group of the, next amino acid molecule with the liberation of water, molecule., , 29. (c) : Four Fe2+ ions of each haemoglobin can, bind with four molecules of O2 and it is carried as, oxyhaemoglobin., 30. (b) : -Helix structure is formed when the chain of, -amino acids coil as a right handed screw because of, the formation of hydrogen bonds between amide groups, of the same peptide chain, i.e., NH group in one unit is, linked to carbonyl oxygen of the fourth unit by hydrogen, bonding. This H-bonding is responsible for holding helix, in a stable position., 31. (b) : Some proteins are also found in D-form., 32. (a) : Peptide bond is formed by the reaction of, — COOH group of one amino acid with the — NH2, group of another amino acid and represented as, , www.neetujee.com, , O, , .., – C – NH –, , O–, –C, , +, , N H–, , As some double bond character is found between C – N, bond, the bond length of C–N in protein should be, smaller than the usual C–N bond., 33. (a) : Starch is also known as amylum which occurs, in all green plants. A molecule of starch (C6H10O5)n is, built of a large number of -glucose rings joined through, oxygen atoms., 34. (d) : Haemoglobin is a globular protein of four, subunits, each subunit having a heme moiety and a, polypeptide chain (Two  and two  chains)., 35. (a), pepsin, 36. (d) : Proteins ,  ,  Polypeptides, proteases, trypsin, ,  , Amino acids, chemotrypsin, , 37. (b), 38. (b) : Diastase is an enzyme that hydrolyses starch, into maltose., 39. (a), 40. (b), 41. (b) : Enzymes being biocatalyst increases the rate, of a chemical reaction by providing alternative lower, activation energy pathways., 42. (b) : Deficiency disease Vitamin, Convulsions, B6, Beri-beri, B1, Cheilosis, B2, Sterility, E, 43. (a) : Vitamin B complex is not a fat soluble vitamin., It is a water soluble vitamin., 44. (d) : Vitamin B and C are water soluble whereas, vitamin A, D, E and K are fat soluble., 45. (c) : Certain organic substances required for, regulating some of the body processes and preventing, certain diseases are called vitamins, which cannot be, synthesised by human body., 46. (b) : Vitamin B12 is chemically named as, cyanocobalamine, having, molecular, formula, C63H88O14N14PCo., 47. (c) : Genetic information flows from, Transcription, Translation, DNA,   ,  RNA ,    Proteins, 48. (d), 49. (b) : Nitrogeneous bases are linked together by, hydrogen bonds., , 50. (b) : Genes are responsible for protein synthesis., 51. (c) : DNA contains two types of nitrogeneous bases, Purine  Adenine (A) and guanine (G), , www.mediit.in

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126, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , Pyrimidine  Cytosine (C) and thymine (T), The purine and pyrimidine bases pair only in certain, combination. Adenine pairs with thymine (A : T) by two, hydrogen bonds and guanine with cytosine (G : C) by, three hydrogen bonds., 52. (c) : The constituents of nucleic acids are, nitrogenous bases, sugar and phosphoric acid. The, sugar present in DNA is D(–)-2-deoxyribose and the, sugar present in RNA is D(–)-ribose. Due to these D(–, )-sugar components, DNA and RNA molecules are, chiral molecules., 2, , O, , 2, , H, , H, , H, , H, H, , OH OH, -D(–)-ribose, , 60. (c) : HO, , O, I, , H, CH2, , I, , C COOH, NH2, , 61. (c) : Thyroxine is an amine hormone and water, soluble hormone containing amino group., , OH, , 62. (b) : Glucagon is a peptide hormone, synthesised by, the -cells of the pancreas., , H, , H, , H, OH, , I, , I, , H, , -D(–)-2-de oxyribose, , 53. (a) : The four bases in mRNA : adenine, cytosine,, guanine and uracil have been shown to act in the form of, triplets; each triplet behaving as a code for the synthesis, of a particular amino acid., 54. (a) : Amount of A = T and that of G = C., 55. (a), 56. (d) : DNA is an example of biopolymer., 57. (a), 58. (c) : Denaturation changes the structure of a protein, and protein loses its activity., 59. (c) : Adrenaline hormone helps to release fatty, acids from fat and glucose from liver glycogen under the, condition of stress. Hence, it is also called ‘flight or fight, hormone’., , 63. (d) : Insulin is a hormone secreted by the pancreas, that lowers blood glucose level by promoting the uptake, of glucose by cells and the conversion of glucose to, glycogen by the liver and skeletal muscle., 64. (a) : It is a hormone secreted from thyroid gland. It, controls various biochemical reactions involving burning, of proteins, carbohydrates, fats to release energy., 65. (c) : Cell membranes are mainly composed of, phospholipids., 66. (b) : Phospholipids may be regarded as derivatives, of glycerol in which two of the hydroxyl groups are, esterified with fatty acids while the third is esterified with, some derivatives of phosphoric acid., 67. (c) : In the lipid metabolism, a molecule of, palmitic acid (C15H31 – COOH) produces 130 adenosine, triphosphate molecules (ATP)., , , , www.neetujee.com, , www.mediit.in

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, , , , , , , , , CHAPTER, , 15, , Polymers, O, , 15.2 Types of Polymerisation Reactions, 1., , (a), (b), (c), (d), 2., , 3., , 4., , (d), , Which of the following is a natural polymer?, cis-1, 4-polyisoprene, poly (Butadiene-styrene), polybutadiene, poly (Butadiene-acrylonitrile), , 5., (NEET 2020), , The polymer that is used as a substitute for wool in, making commercial fibres is, (a) melamine, (b) nylon-6, 6, (c) polyacrylonitrile, (d) buna-N., (Odisha NEET 2019), Regarding cross-linked or network polymers, which, of the following statements is incorrect?, , 6., , 7., , (a) They contain covalent bonds between various, linear polymer chains., (b) They are formed from bi- and tri-functional, monomers., (c) Examples are bakelite and melamine., (d) They contain strong covalent bonds in their, polymer chains., (NEET 2018), Which one of the following structures represents, , (b), , (c), , NH2, , www.neetujee.com, , Caprolactam is used for the manufacture of, (a) teflon, (b) terylene, (c) nylon 6, 6, (d) nylon 6., , CH CH2 )n, , C, , Cl, H O, , O, , (c) ( N (CH2 )6 N C (CH2 )4 C )n, OH, , OH, CH2, , CH 2, n, , 8., , H2 H H2 H, C C C C, CH3, , COOH 6, , (2015), , Which one of the following is an example of, thermosetting polymer?, , H, , NH2 66, , Cl 6, , n, , O, , (b) ( CH2 CH )n, , CH3 66, , H2, H2 H, C CH C C, , 2, , Cl, , H 2 H H2 H, C C C C, NH2, , H, N ( CH 2 ) 6 NH, , H2, CC, , (NEET-II 2016), Natural rubber has, (a) alternate cis- and trans-configuration, (b) random cis- and trans-configuration, (c) all cis-configuration, (d) all trans-configuration., (NEET-I 2016), , (d), , H 2 H H2 H, C C C C, NH2, , C, H2, , (a) ( CH2, , nylon 6, 6 polymer?, (a), , C, , 9., , (2014), , Which of the following organic compounds, polymerizes to form the polyester dacron?, (a) Propylene and para HO—(C6H4)—OH, (b) Benzoic acid and ethanol, (c) Terephthalic acid and ethylene glycol, (d) Benzoic acid and para HO—(C6H4)—OH, (2014), Nylon is an example of, (a) polyamide, (b) polythene, (c) polyester, (d) polysaccharide., (NEET 2013), , www.mediit.in

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128, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 10. Which is the monomer of neoprene in the following?, , C — CH, , (a) CH2, , CH2, , Cl, (b) CH2, , CH—C CH, , (c) CH2, , CH—CH, , CH2, , (d) CH2, , C — CH, , CH2, , (NEET 2013, 2003), , CH3, 11. Which one of the following is not a condensation, polymer?, (a) Melamine, (b) Glyptal, (c) Dacron, (d) Neoprene, (2012), 12. Which of the following statements is false?, (a) Artificial silk is derived from cellulose., (b) Nylon-6,6 is an example of elastomer., (c) The repeat unit in natural rubber is isoprene., (d) Both starch and cellulose are polymers of, 13. Of the following which one is classified as polyester, polymer?, (a) Terylene, (b) Bakelite, (c) Melamine, (d) Nylon-6,6, (2011), 14. Which of the following structures represents, neoprene polymer?, (a) [ CH2 C, , CH CH2 n, , Cl, , [ NH(CH2)6NHCO(CH2)4CO ]n is a, , (a) homopolymer, (b) copolymer, (c) addition polymer, (d) thermosetting polymer., 19. The monomer of the polymer, , CN, (c) [ CH2 CH n, [ CH CH2 n, , (2010), , C6H5, 15. Structures of some common polymers are given., Which one is not correctly presented?, (a) Neoprene- CH2 C CH CH2 CH2, n, , Cl, , (b) Terylene -, , OC, , (2006), , CH3, , CH3, is, CH2 – C– CH2 – C, CH3, CH3, CH3, (a) H2C C, CH3, (b) CH3CH, , CHCH3, , (c) CH3CH, , CH2, C(CH3)2, , (2005), , 20. Which one of the following is a chain growth, polymer?, (a) Starch, (b) Nucleic acid, (c) Polystyrene, (d) Protein, (2004), , (b) [ CH2 CH n, , COOCH2 CH2 O, , (c) Nylon 6,6 - NH(CH2) 6 NHCO(CH2 )4 CO, , www.neetujee.com, , 18., , (d) (CH3)2C, , Cl, , (d) Teflon -, , 17. Which one of the following polymers is prepared by, condensation polymerisation?, (a) Teflon, (b) Natural rubber, (c) Styrene, (d) Nylon-6,6, (2007), , (2012), , glucose., , (d), , 16. Which one of the following statements is not true?, (a) Buna-S is a copolymer of butadiene and styrene., (b) Natural rubber is a 1,4-polymer of isoprene., (c) In vulcanization, the formation of sulphur, bridges between different chains make rubber, harder and stronger., (d) Natural rubber has the trans-configuration at, every double bond., (2008), , n, , 21. Acrilan is a hard, horny and a high melting material., Which one of the following represents its structure?, (a), , CH3, (b) –CH2 – C–, COOCH3 n, (c), , –CH2 – CH–, COOC2H5 n, –CH 2 –CH–, , n, , (2009), , –CH2–CH–, CN n, , (d), , Cl, , (2003), , n, , www.mediit.in

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Polymers, , 129, , CH3, 22. Monomer of –C –CH2– is, , CH3, (a) 2-methylpropene, (c) propylene, , n, , (b) styrene, (d) ethene., , (2002), , 28. Which one of the following is used to make ‘nonstick’ cookware?, (a) Polyethylene terephthalate, (b) Polytetrafluoroethylene, (c) PVC, (d) Polystyrene, (1997), , 23. Which of the following is not correctly matched?, (a) Neoprene :, , –CH2 –C CH –CH2–, Cl, n, (b) Nylon-6,6 :, O, –NH–(CH2)6–NH– CO – (CH2 )4– C – O–n, , (c) Terylene :, , O, , O, , –OCH2 –CH2– C–, , –C, , n, , (d) PMMA : –CH2–C–., , COOCH3 n, , (2001), , CF2 is monomer of, , (a) teflon, (c) polythene, , 29. The bakelite is prepared by the reaction between, (a) phenol and formaldehyde, (b) tetramethylene glycol, (c) urea and formaldehyde, (d) ethylene glycol., (1995), , 15.4 Biodegradable Polymers, 30. The biodegradable polymer is, (a) buna-S, (b) nylon-6,6, (c) nylon-2-nylon 6, (d) nylon-6. (NEET 2019), , CH3, , 24. CF2, , 27. Terylene is a condensation polymer of ethylene, glycol and, (a) salicylic acid, (b) phthalic acid, (c) benzoic acid, (d) terephthalic acid., (1999), , (b) orlon, (d) nylon-6., , 31. Which one of the following sets forms the, biodegradable polymer?, (a) CH2 CH — CN and, CH2 CH — CH CH2, (b) H2N — CH2 — COOH, and H2N — (CH2)5 — COOH, (c) HO — CH2 — CH2 — OH and, , (2000), , 25. Which compound forms linear polymer due to Hbond?, (a) H2O, (b) NH3, (c) HF, (d) HCl, (2000), 26. Natural rubber is a polymer of, (a) styrene, (b) ethyne, (c) butadiene, (d) isoprene., , HOOC, CH, , (d), (1999), , COOH, , CH2, , CH2 and, , CH — CH, , (Mains 2012), , CH2, , ANSWER KEY, , 1., 11., 21., 31., , (a), (d), (a), (b), , www.neetujee.com, , 2., 12., 22., , (c), (b), (a), , 3., 13., 23., , (d), (a), (c), , 4., 14., 24., , (d), (a), (a), , 5., 15., 25., , (c), (a), (c), , 6., 16., 26., , (d), (d), (d), , 7., 17., 27., , (d), (d), (d), , 8., 18., 28., , (c), (b), (b), , 9., 19., 29., , (a), (a), (a), , 10., 20., 30., , (a), (c), (c), , www.mediit.in

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Polymers, , 131, , 20. (c) : Chain-growth polymers involve a series of, reactions each of which consume a reactive particle and, produces another similar one. The reactive particles, may be free radicals or ions (cation or anion) to which, monomers get added by a chain reaction. It is an, important reaction of alkenes and conjugated dienes or, indeed of all kinds of compounds that contains C – C, double bonds., CH2CH3, , CH, , CH2 CH 2, , Fe2O3/Cr2O3, , AlCl3, , 650°C, , CH2, , 24. (a) : n(CF2, , CF2)  CF2 CF2, , 25. (c) : H-F – – – H-F– – – H-F– – – H-F, Dotted lines represent hydrogen bond between HF, molecules and hence, it is a linear polymer. Due to high, electronegativity value of ‘F’ atom, it forms effective, hydrogen bonding., 26. (d) :, , Isoprene, , 21. (a) : Acrilan is an addition polymer of acrylonitrile., CH CN, , Polyisoprene, , Polyisoprene is the natural rubber, which is the polymer, of isoprene., , Polystyrene, , n CH2, , n, , Polytetrafluoroethylene, (Teflon), , 27. (d) : n( HO—CH2—CH2—OH) +, Ethylene glycol, , CH2 CH, CN n, CH3, , 22. (a) : The monomer of, , C CH, CH3, , is H3, , 2, , n, , CH2, CH3, , (2-methylpropene), , 23. (c) : Terylene is an example of condensation, polymer and formed by the condensation of terephthalic, and ethylene glycol., , Terylene is the condensation polymer of ethylene glycol, and terephthalic acid., 28. (b) : Polytetrafluoroethylene or teflon is a tough, material, resistance to heat and bad conductor of, electricity. It is used for coating the cookware to make, them non-sticky., 29. (a) : Phenol and formaldehyde undergo, condensation polymerisation under two different, conditions to give a cross linked polymer called bakelite., 30. (c) : Nylon-2-nylon-6 is a biodegradable polymer., 31. (b) : Nylon-2-nylon-6 is an alternating polyamide, copolymerofglycine(H2NCH2COOH) andaminocaproic, acid (H2N(CH2)5COOH) and is biodegradable., , , , www.neetujee.com, , www.mediit.in

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, , , , , , , CHAPTER, , 16, , Chemistry in, Everyday Life, (c) phenol and iodine, (d) terpineol and bithionol., (Karnataka NEET 2013), , 16.3 Therapeutic Action of Different Classes, of Drugs, 1., , Among the following, the narrow spectrum, antibiotic is, (a) chloramphenicol, (b) penicillin G, (c) ampicillin, (d) amoxycillin., (NEET 2019), , 2., , Mixture of chloroxylenol and terpineol acts as, (a) antiseptic, (b) antipyretic, (c) antibiotic, (d) analgesic., (NEET 2017), , 3., , Which of the following is an analgesic?, (a) Streptomycin, (b) Chloromycetin, (c) Novalgin, (d) Penicillin, (NEET-I 2016), , 4., , 5., , 6., , 7., , Chloroamphenicol is an, (a) antifertility drug, (b) antihistamine, (c) antiseptic and disinfectant, (d) antibiotic-broad spectrum., , (Mains 2012), , 8., , Which one of the following is employed as, Antihistamine?, (a) Chloramphenicol, (b) Diphenylhydramine, (c) Norethindrone, (d) Omeprazole, (2011), , 9., , Which one of the following is employed as a, tranquilizer drug?, (a) Promethazine, (b) Valium, (c) Naproxen, (d) Mifepriston, (2010), , 10., , Which one of the following is employed as a tranquilizer?, (a) Naproxen, (b) Tetracycline, (c) Chlorpheniramine (d) Equanil, (2009), , Antiseptics and disinfectants either kill or prevent, growth of microrganisms. Identify which of the, following statements is not true., (a) Dilute solutions of boric acid and hydrogen, peroxide are strong antiseptics., (b) Disinfectants harm the living tissues., (c) A 0.2% solution of phenol is an antiseptic while, 1% solution acts as a disinfectant., (d) Chlorine and iodine are used as strong, disinfectants., (NEET 2013), , 11., , Chloropicrin is obtained by the reaction of, (a) steam on carbon tetrachloride, (b) nitric acid on chlorobenzene, (c) chlorine on picric acid, (d) nitric acid on chloroform., (2004), , 12., , Aspirin is an acetylation product of, (a) m-hydroxybenzoic acid, (b) o-dihydroxybenzene, (c) o-hydroxybenzoic acid, (d) p-dihydroxybenzene., , Dettol is the mixture of, (a) chloroxylenol and bithionol, (b) chloroxylenol and terpineol, , 13., , Bithional is generally added to the soaps as an, additive to function as a/an, (a) buffering agent, (b) antiseptic, (c) softener, (d) dryer., (2015, Cancelled), , www.neetujee.com, , (1998), , Which of the following can possibly be used as, analgesic without causing addiction and mood, modification?, , www.mediit.in

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Chemistry in Everyday Life, (a), (b), (c), (d), 14., , 15., , 133, , Diazepam, Tetrahydrocatinol, Morphine, N-Acetyl-para-aminophenol., , (1997), , Which one of the following statements is not true?, (a) Ampicillin is a natural antibiotic., (b) Aspirin is both analgesic and antipyretic., (c) Sulphadiazine is a synthetic antibacterial drug., (d) Some disinfectants can be used as antiseptics., (1994), Diazo coupling is useful to prepare some, (a) pesticides, (b) dyes, (c) proteins, (d) vitamins., , (1994), , 16.4 Chemicals in Foods, 16., , The artificial sweetner stable at cooking temperature, and does not provide calories is, (a) saccharin, (b) aspartame, (c) sucralose, (d) alitame., (Odisha NEET 2019), , 17. Artificial sweetner which is stable under cold, conditions only is, (a) saccharine, (b) sucralose, (c) aspartame, (d) alitame., (2014), , 16.5 Cleansing Agents, 18. Which of the following is a cationic detergent?, (a) Sodium lauryl sulphate, (b) Sodium stearate, (c) Cetyltrimethyl ammonium bromide, (d) Sodium dodecylbenzene sulphonate, (NEET 2020), 19. Which of the following forms cationic micelles, above certain concentration?, (a) Sodium dodecyl sulphate, (b) Sodium acetate, (c) Urea, (d) Cetyltrimethylammonium bromide, (2004), , ANSWER KEY, , 1., 11., , (b), (d), , 2., 12., , (a), (c), , 3., 13., , (c), (d), , 4., 14., , (b), (a), , 5., 15., , (a), (b), , 6., 16., , (b), (c), , 7., 17., , (d), (c), , 8., 18., , (b), (c), , 9., 19., , (b), (d), , 10., , (d), , Hints & Explanations, 1. (b) : Penicillin G has a narrow spectrum., Chloramphenicol is a broad spectrum antibiotic., Ampicillin and amoxycillin are synthetic modifications, of penicillins. These have broad spectrum., 2. (a) : Dettol which is a well known antiseptic is a, mixture of chloroxylenol and -terpineol in a suitable, solvent., , 8. (b) : Diphenylhydramine, antihistamine drug., 9., , is, , employed, , (b) : Valium is a tranquilizer., , 10. (d) : Equanil is used for the treatment of stress, mild, and severe mental diseases i.e., as a tranquilizer., 11. (d) : When chloroform is treated with concentrated, nitric acid, its hydrogen is replaced by nitro group., , 3. (c) : Streptomycin, chloromycetin and penicillin, are antibiotics while novalgin is an analgesic., , CHCl3 + HONO2  CNO2Cl3 + H2O, , 4., , 12. (c) : Aspirin is acetyl salicylic acid, which, is formed by acetylation of o-hydroxybenzoic acid., OCOCH3, OH, , (b), , 5. (a) : Dilute solutions of boric acid and hydrogen, peroxide are weak antiseptics., 6., , (b) : Dettol is the mixture of chloroxylenol and, , -terpineol., 7., , (d), , www.neetujee.com, , as, , Chloropicrin, , COOH CH3COCl, o-Hydroxybenzoic acid, , COOH, Aspirin, , www.mediit.in

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134, , NEET-AIPMT Chapterwise Topicwise Solutions Chemistry, , 13. (d) : N-Acetyl-para-aminophenol (or paracetamol), is an antipyretic which can also be used as an analgesic to, relieve pain without addition and mood modification., , 16. (c) : Sucralose is trichloro derivative of sucrose. Its, appearance and taste is like sugar. It is stable at cooking, temperature and it does not provide calories., , 14. (a) : Ampicillin is a modification of penicillin and, thus is a synthetic antibiotic., , 17. (c) : Aspartame is stable under cold conditions and, unstable at cooking temperature., , 15. (b) : Azo dyes are derived by coupling of a, phenol adsorbed on the surface of a fabric with a, diazonium salt. Dyes can be prepared by diazo coupling., For example,, , 18. (c) : Cetyltrimethyl ammonium bromide is a, cationic detergent., CH3, +, CH3 (CH2)15 N CH3 Br–, CH3, , COOH, N, , N, , Methyl red, , N(CH3)2, , 19. (d) : Cetyltrimethylammonium, popular cationic detergent., , bromide, , is, , a, , , , www.neetujee.com, , www.mediit.in